Problem 1
=========

If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.


Problem 2
=========

Each new term in the Fibonacci sequence is generated by adding the
previous two terms. By starting with 1 and 2, the first 10 terms will be:

                  1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

Find the sum of all the even-valued terms in the sequence which do not
exceed four million.


Problem 3
=========

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143?


Problem 4
=========

A palindromic number reads the same both ways. The largest palindrome made
from the product of two 2-digit numbers is 9009 = 91 * 99.

Find the largest palindrome made from the product of two 3-digit numbers.


Problem 5
=========

2520 is the smallest number that can be divided by each of the numbers
from 1 to 10 without any remainder.

What is the smallest number that is evenly divisible by all of the numbers
from 1 to 20?


Problem 6
=========

The sum of the squares of the first ten natural numbers is,
                       1^2 + 2^2 + ... + 10^2 = 385

The square of the sum of the first ten natural numbers is,
                    (1 + 2 + ... + 10)^2 = 55^2 = 3025

Hence the difference between the sum of the squares of the first ten
natural numbers and the square of the sum is 3025 - 385 = 2640.

Find the difference between the sum of the squares of the first one
hundred natural numbers and the square of the sum.


Problem 7
=========

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see
that the 6th prime is 13.

What is the 10001st prime number?


Problem 8
=========

Find the greatest product of thirteen consecutive digits in the 1000-digit
number.

            73167176531330624919225119674426574742355349194934
            96983520312774506326239578318016984801869478851843
            85861560789112949495459501737958331952853208805511
            12540698747158523863050715693290963295227443043557
            66896648950445244523161731856403098711121722383113
            62229893423380308135336276614282806444486645238749
            30358907296290491560440772390713810515859307960866
            70172427121883998797908792274921901699720888093776
            65727333001053367881220235421809751254540594752243
            52584907711670556013604839586446706324415722155397
            53697817977846174064955149290862569321978468622482
            83972241375657056057490261407972968652414535100474
            82166370484403199890008895243450658541227588666881
            16427171479924442928230863465674813919123162824586
            17866458359124566529476545682848912883142607690042
            24219022671055626321111109370544217506941658960408
            07198403850962455444362981230987879927244284909188
            84580156166097919133875499200524063689912560717606
            05886116467109405077541002256983155200055935729725
            71636269561882670428252483600823257530420752963450


Problem 9
=========

A Pythagorean triplet is a set of three natural numbers, a < b < c, for
which,
                             a^2 + b^2 = c^2

For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.


Problem 10
==========

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.


Problem 11
==========

In the 20 * 20 grid below, four numbers along a diagonal line have been
marked in red.

       08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
       49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
       81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
       52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
       22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
       24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
       32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
       67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
       24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
       21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
       78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
       16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
       86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
       19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
       04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
       88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
       04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
       20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
       20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
       01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 * 63 * 78 * 14 = 1788696.

What is the greatest product of four adjacent numbers in any direction
(up, down, left, right, or diagonally) in the 20 * 20 grid?


Problem 12
==========

The sequence of triangle numbers is generated by adding the natural
numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 =
28. The first ten terms would be:

                 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

   1: 1
   3: 1,3
   6: 1,2,3,6
  10: 1,2,5,10
  15: 1,3,5,15
  21: 1,3,7,21
  28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five
divisors.

What is the value of the first triangle number to have over five hundred
divisors?


Problem 13
==========

Work out the first ten digits of the sum of the following one-hundred
50-digit numbers.

            37107287533902102798797998220837590246510135740250
            46376937677490009712648124896970078050417018260538
            74324986199524741059474233309513058123726617309629
            91942213363574161572522430563301811072406154908250
            23067588207539346171171980310421047513778063246676
            89261670696623633820136378418383684178734361726757
            28112879812849979408065481931592621691275889832738
            44274228917432520321923589422876796487670272189318
            47451445736001306439091167216856844588711603153276
            70386486105843025439939619828917593665686757934951
            62176457141856560629502157223196586755079324193331
            64906352462741904929101432445813822663347944758178
            92575867718337217661963751590579239728245598838407
            58203565325359399008402633568948830189458628227828
            80181199384826282014278194139940567587151170094390
            35398664372827112653829987240784473053190104293586
            86515506006295864861532075273371959191420517255829
            71693888707715466499115593487603532921714970056938
            54370070576826684624621495650076471787294438377604
            53282654108756828443191190634694037855217779295145
            36123272525000296071075082563815656710885258350721
            45876576172410976447339110607218265236877223636045
            17423706905851860660448207621209813287860733969412
            81142660418086830619328460811191061556940512689692
            51934325451728388641918047049293215058642563049483
            62467221648435076201727918039944693004732956340691
            15732444386908125794514089057706229429197107928209
            55037687525678773091862540744969844508330393682126
            18336384825330154686196124348767681297534375946515
            80386287592878490201521685554828717201219257766954
            78182833757993103614740356856449095527097864797581
            16726320100436897842553539920931837441497806860984
            48403098129077791799088218795327364475675590848030
            87086987551392711854517078544161852424320693150332
            59959406895756536782107074926966537676326235447210
            69793950679652694742597709739166693763042633987085
            41052684708299085211399427365734116182760315001271
            65378607361501080857009149939512557028198746004375
            35829035317434717326932123578154982629742552737307
            94953759765105305946966067683156574377167401875275
            88902802571733229619176668713819931811048770190271
            25267680276078003013678680992525463401061632866526
            36270218540497705585629946580636237993140746255962
            24074486908231174977792365466257246923322810917141
            91430288197103288597806669760892938638285025333403
            34413065578016127815921815005561868836468420090470
            23053081172816430487623791969842487255036638784583
            11487696932154902810424020138335124462181441773470
            63783299490636259666498587618221225225512486764533
            67720186971698544312419572409913959008952310058822
            95548255300263520781532296796249481641953868218774
            76085327132285723110424803456124867697064507995236
            37774242535411291684276865538926205024910326572967
            23701913275725675285653248258265463092207058596522
            29798860272258331913126375147341994889534765745501
            18495701454879288984856827726077713721403798879715
            38298203783031473527721580348144513491373226651381
            34829543829199918180278916522431027392251122869539
            40957953066405232632538044100059654939159879593635
            29746152185502371307642255121183693803580388584903
            41698116222072977186158236678424689157993532961922
            62467957194401269043877107275048102390895523597457
            23189706772547915061505504953922979530901129967519
            86188088225875314529584099251203829009407770775672
            11306739708304724483816533873502340845647058077308
            82959174767140363198008187129011875491310547126581
            97623331044818386269515456334926366572897563400500
            42846280183517070527831839425882145521227251250327
            55121603546981200581762165212827652751691296897789
            32238195734329339946437501907836945765883352399886
            75506164965184775180738168837861091527357929701337
            62177842752192623401942399639168044983993173312731
            32924185707147349566916674687634660915035914677504
            99518671430235219628894890102423325116913619626622
            73267460800591547471830798392868535206946944540724
            76841822524674417161514036427982273348055556214818
            97142617910342598647204516893989422179826088076852
            87783646182799346313767754307809363333018982642090
            10848802521674670883215120185883543223812876952786
            71329612474782464538636993009049310363619763878039
            62184073572399794223406235393808339651327408011116
            66627891981488087797941876876144230030984490851411
            60661826293682836764744779239180335110989069790714
            85786944089552990653640447425576083659976645795096
            66024396409905389607120198219976047599490197230297
            64913982680032973156037120041377903785566085089252
            16730939319872750275468906903707539413042652315011
            94809377245048795150954100921645863754710598436791
            78639167021187492431995700641917969777599028300699
            15368713711936614952811305876380278410754449733078
            40789923115535562561142322423255033685442488917353
            44889911501440648020369068063960672322193204149535
            41503128880339536053299340368006977710650566631954
            81234880673210146739058568557934581403627822703280
            82616570773948327592232845941706525094512325230608
            22918802058777319719839450180888072429661980811197
            77158542502016545090413245809786882778948721859617
            72107838435069186155435662884062257473692284509516
            20849603980134001723930671666823555245252804609722
            53503534226472524250874054075591789781264330331690


Problem 14
==========

The following iterative sequence is defined for the set of positive
integers:

n->n/2 (n is even)
n->3n+1 (n is odd)

Using the rule above and starting with 13, we generate the following
sequence:
                  13->40->20->10->5->16->8->4->2->1

It can be seen that this sequence (starting at 13 and finishing at 1)
contains 10 terms. Although it has not been proved yet (Collatz Problem),
it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.


Problem 15
==========

Starting in the top left corner of a 2 * 2 grid, there are 6 routes
(without backtracking) to the bottom right corner.

How many routes are there through a 20 * 20 grid?


Problem 16
==========

2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number 2^1000?


Problem 17
==========

If the numbers 1 to 5 are written out in words: one, two, three, four,
five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written
out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
forty-two) contains 23 letters and 115 (one hundred and fifteen) contains
20 letters. The use of "and" when writing out numbers is in compliance
with British usage.


Problem 18
==========

By starting at the top of the triangle below and moving to adjacent
numbers on the row below, the maximum total from top to bottom is 23.

                                    3
                                   7 4
                                  2 4 6
                                 8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

                                    75
                                  95 64
                                 17 47 82
                               18 35 87 10
                              20 04 82 47 65
                            19 01 23 75 03 34
                           88 02 77 73 07 63 67
                         99 65 04 28 06 16 70 92
                        41 41 26 56 83 40 80 70 33
                      41 48 72 33 47 32 37 16 94 29
                     53 71 44 65 25 43 91 52 97 51 14
                   70 11 33 28 77 73 17 78 39 68 17 57
                  91 71 52 38 17 14 91 43 58 50 27 29 48
                63 66 04 68 89 53 67 30 73 16 69 87 40 31
               04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem
   by trying every route. However, Problem 67, is the same challenge with
a triangle containing one-hundred rows; it cannot be solved by brute
force, and requires a clever method! ;o)


Problem 19
==========

You are given the following information, but you may prefer to do some
research for yourself.

  * 1 Jan 1900 was a Monday.
  * Thirty days has September,
    April, June and November.
    All the rest have thirty-one,
    Saving February alone,
    Which has twenty-eight, rain or shine.
    And on leap years, twenty-nine.
  * A leap year occurs on any year evenly divisible by 4, but not on a
    century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth
century (1 Jan 1901 to 31 Dec 2000)?


Problem 20
==========

n! means n * (n - 1) * ... * 3 * 2 * 1

Find the sum of the digits in the number 100!


Problem 21
==========

Let d(n) be defined as the sum of proper divisors of n (numbers less than
n which divide evenly into n).
If d(a) = b and d(b) = a, where a =/= b, then a and b are an amicable pair
and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22,
44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1,
2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.


Problem 22
==========

Using names.txt, a 46K text file containing over five-thousand first names,
begin by sorting it into alphabetical order. Then working out the
alphabetical value for each name, multiply this value by its alphabetical
position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which
is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So,
COLIN would obtain a score of 938 * 53 = 49714.

What is the total of all the name scores in the file?


Problem 23
==========

A perfect number is a number for which the sum of its proper divisors is
exactly equal to the number. For example, the sum of the proper divisors
of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect
number.

A number whose proper divisors are less than the number is called
deficient and a number whose proper divisors exceed the number is called
abundant.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the
smallest number that can be written as the sum of two abundant numbers is
24. By mathematical analysis, it can be shown that all integers greater
than 28123 can be written as the sum of two abundant numbers. However,
this upper limit cannot be reduced any further by analysis even though it
is known that the greatest number that cannot be expressed as the sum of
two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the
sum of two abundant numbers.


Problem 24
==========

A permutation is an ordered arrangement of objects. For example, 3124 is
one possible permutation of the digits 1, 2, 3 and 4. If all of the
permutations are listed numerically or alphabetically, we call it
lexicographic order. The lexicographic permutations of 0, 1 and 2 are:

                    012   021   102   120   201   210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3,
4, 5, 6, 7, 8 and 9?


Problem 25
==========

The Fibonacci sequence is defined by the recurrence relation:

  F[n] = F[n[1]] + F[n[2]], where F[1] = 1 and F[2] = 1.

Hence the first 12 terms will be:

  F[1] = 1
  F[2] = 1
  F[3] = 2
  F[4] = 3
  F[5] = 5
  F[6] = 8
  F[7] = 13
  F[8] = 21
  F[9] = 34
  F[10] = 55
  F[11] = 89
  F[12] = 144

The 12th term, F[12], is the first term to contain three digits.

What is the first term in the Fibonacci sequence to contain 1000 digits?


Problem 26
==========

A unit fraction contains 1 in the numerator. The decimal representation of
the unit fractions with denominators 2 to 10 are given:

   1/2  =  0.5
   1/3  =  0.(3)
   1/4  =  0.25
   1/5  =  0.2
   1/6  =  0.1(6)
   1/7  =  0.(142857)
   1/8  =  0.125
   1/9  =  0.(1)
  1/10  =  0.1

Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can
be seen that ^1/[7] has a 6-digit recurring cycle.

Find the value of d < 1000 for which ^1/[d] contains the longest recurring
cycle in its decimal fraction part.


Problem 27
==========

Euler published the remarkable quadratic formula:

                               n^2 + n + 41

It turns out that the formula will produce 40 primes for the consecutive
values n = 0 to 39. However, when n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41
is divisible by 41, and certainly when n = 41, 41^2 + 41 + 41 is clearly
divisible by 41.

Using computers, the incredible formula  n^2 - 79n + 1601 was discovered,
which produces 80 primes for the consecutive values n = 0 to 79. The
product of the coefficients, 79 and 1601, is 126479.

Considering quadratics of the form:

  n^2 + an + b, where |a| < 1000 and |b| < 1000

                              where |n| is the modulus/absolute value of n
                                               e.g. |11| = 11 and |-4| = 4

Find the product of the coefficients, a and b, for the quadratic
expression that produces the maximum number of primes for consecutive
values of n, starting with n = 0.


Problem 28
==========

Starting with the number 1 and moving to the right in a clockwise
direction a 5 by 5 spiral is formed as follows:

                              21 22 23 24 25
                              20  7  8  9 10
                              19  6  1  2 11
                              18  5  4  3 12
                              17 16 15 14 13

It can be verified that the sum of both diagonals is 101.

What is the sum of both diagonals in a 1001 by 1001 spiral formed in the
same way?


Problem 29
==========

Consider all integer combinations of a^b for 2 a 5 and 2 b 5:

  2^2=4, 2^3=8, 2^4=16, 2^5=32
  3^2=9, 3^3=27, 3^4=81, 3^5=243
  4^2=16, 4^3=64, 4^4=256, 4^5=1024
  5^2=25, 5^3=125, 5^4=625, 5^5=3125

If they are then placed in numerical order, with any repeats removed, we
get the following sequence of 15 distinct terms:

     4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

How many distinct terms are in the sequence generated by a^b for
2 <= a <= 100 and 2 <= b <= 100?


Problem 30
==========

Surprisingly there are only three numbers that can be written as the sum
of fourth powers of their digits:

  1634 = 1^4 + 6^4 + 3^4 + 4^4
  8208 = 8^4 + 2^4 + 0^4 + 8^4
  9474 = 9^4 + 4^4 + 7^4 + 4^4

As 1 = 1^4 is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth
powers of their digits.


Problem 31
==========

In England the currency is made up of pound, -L-, and pence, p, and there
are eight coins in general circulation:

  1p, 2p, 5p, 10p, 20p, 50p, -L-1 (100p) and -L-2 (200p).

It is possible to make -L-2 in the following way:

  1 * -L-1 + 1 * 50p + 2 * 20p + 1 * 5p + 1 * 2p + 3 * 1p

How many different ways can -L-2 be made using any number of coins?


Problem 32
==========

We shall say that an n-digit number is pandigital if it makes use of all
the digits 1 to n exactly once; for example, the 5-digit number, 15234,
is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 * 186 = 7254, containing
multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product
identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to
only include it once in your sum.


Problem 33
==========

The fraction 49/98 is a curious fraction, as an inexperienced
mathematician in attempting to simplify it may incorrectly believe that
49/98 = 4/8, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, 30/50 = 3/5, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less
than one in value, and containing two digits in the numerator and
denominator.

If the product of these four fractions is given in its lowest common
terms, find the value of the denominator.


Problem 34
==========

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.

Find the sum of all numbers which are equal to the sum of the factorial of
their digits.

Note: as 1! = 1 and 2! = 2 are not sums they are not included.


Problem 35
==========

The number, 197, is called a circular prime because all rotations of the
digits: 197, 971, and 719, are themselves prime.

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37,
71, 73, 79, and 97.

How many circular primes are there below one million?


Problem 36
==========

The decimal number, 585 = 1001001001[2] (binary), is palindromic in both
bases.

Find the sum of all numbers, less than one million, which are palindromic
in base 10 and base 2.

(Please note that the palindromic number, in either base, may not include
leading zeros.)


Problem 37
==========

The number 3797 has an interesting property. Being prime itself, it is
possible to continuously remove digits from left to right, and remain
prime at each stage: 3797, 797, 97, and 7. Similarly we can work from
right to left: 3797, 379, 37, and 3.

Find the sum of the only eleven primes that are both truncatable from left
to right and right to left.

NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.


Problem 38
==========

Take the number 192 and multiply it by each of 1, 2, and 3:

  192 * 1 = 192
  192 * 2 = 384
  192 * 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We
will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4,
and 5, giving the pandigital, 918273645, which is the concatenated product
of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as
the concatenated product of an integer with (1,2, ... , n) where n > 1?


Problem 39
==========

If p is the perimeter of a right angle triangle with integral length
sides, {a,b,c}, there are exactly three solutions for p = 120.

                    {20,48,52}, {24,45,51}, {30,40,50}

For which value of p < 1000, is the number of solutions maximised?


Problem 40
==========

An irrational decimal fraction is created by concatenating the positive
integers:

                  0.123456789101112131415161718192021...
                               ^

It can be seen that the 12th digit of the fractional part is 1.

If d[n] represents the n-th digit of the fractional part, find the value
of the following expression.

    d[1] * d[10] * d[100] * d[1000] * d[10000] * d[100000] * d[1000000]


Problem 41
==========

We shall say that an n-digit number is pandigital if it makes use of all
the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital
and is also prime.

What is the largest n-digit pandigital prime that exists?


Problem 42
==========

The n-th term of the sequence of triangle numbers is given by, t[n] =
1/2n(n+1); so the first ten triangle numbers are:

                 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

By converting each letter in a word to a number corresponding to its
alphabetical position and adding these values we form a word value. For
example, the word value for SKY is 19 + 11 + 25 = 55 = t[10]. If the word
value is a triangle number then we shall call the word a triangle word.

Using words.txt, a 16K text file containing nearly two-thousand common
English words, how many are triangle words?


Problem 43
==========

The number, 1406357289, is a 0 to 9 pandigital number because it is made
up of each of the digits 0 to 9 in some order, but it also has a rather
interesting sub-string divisibility property.

Let d[1] be the 1st digit, d[2] be the 2nd digit, and so on. In this
way, we note the following:

  * d[2]d[3]d[4]=406 is divisible by 2
  * d[3]d[4]d[5]=063 is divisible by 3
  * d[4]d[5]d[6]=635 is divisible by 5
  * d[5]d[6]d[7]=357 is divisible by 7
  * d[6]d[7]d[8]=572 is divisible by 11
  * d[7]d[8]d[9]=728 is divisible by 13
  * d[8]d[9]d[10]=289 is divisible by 17

Find the sum of all 0 to 9 pandigital numbers with this property.


Problem 44
==========

Pentagonal numbers are generated by the formula, P[n]=n(3n-1)/2. The first
ten pentagonal numbers are:

               1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P[4] + P[7] = 22 + 70 = 92 = P[8]. However, their
difference, 70 - 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, P[j] and P[k], for which their sum
and difference is pentagonal and D = |P[k] - P[j]| is minimised; what is
the value of D?


Problem 45
==========

Triangle, pentagonal, and hexagonal numbers are generated by the following
formulae:

Triangle     T[n]=n(n+1)/2   1, 3, 6, 10, 15, ...
Pentagonal   P[n]=n(3n-1)/2  1, 5, 12, 22, 35, ...
Hexagonal    H[n]=n(2n-1)    1, 6, 15, 28, 45, ...

It can be verified that T[285] = P[165] = H[143] = 40755.

Find the next triangle number that is also pentagonal and hexagonal.


Problem 46
==========

It was proposed by Christian Goldbach that every odd composite number can
be written as the sum of a prime and twice a square.

9 = 7 + 2 * 1^2
15 = 7 + 2 * 2^2
21 = 3 + 2 * 3^2
25 = 7 + 2 * 3^2
27 = 19 + 2 * 2^2
33 = 31 + 2 * 1^2

It turns out that the conjecture was false.

What is the smallest odd composite that cannot be written as the sum of a
prime and twice a square?


Problem 47
==========

The first two consecutive numbers to have two distinct prime factors are:

14 = 2 * 7
15 = 3 * 5

The first three consecutive numbers to have three distinct prime factors
are:

644 = 2^2 * 7 * 23
645 = 3 * 5 * 43
646 = 2 * 17 * 19.

Find the first four consecutive integers to have four distinct primes
factors. What is the first of these numbers?


Problem 48
==========

The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.

Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.


Problem 49
==========

The arithmetic sequence, 1487, 4817, 8147, in which each of the terms
increases by 3330, is unusual in two ways: (i) each of the three terms are
prime, and, (ii) each of the 4-digit numbers are permutations of one
another.

There are no arithmetic sequences made up of three 1-, 2-, or 3-digit
primes, exhibiting this property, but there is one other 4-digit
increasing sequence.

What 12-digit number do you form by concatenating the three terms in this
sequence?


Problem 50
==========

The prime 41, can be written as the sum of six consecutive primes:

                       41 = 2 + 3 + 5 + 7 + 11 + 13

This is the longest sum of consecutive primes that adds to a prime below
one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a
prime, contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of the most
consecutive primes?


Problem 51
==========

By replacing the 1st digit of *57, it turns out that six of the possible
values: 157, 257, 457, 557, 757, and 857, are all prime.

By replacing the 3rd and 4th digits of 56**3 with the same digit, this
5-digit number is the first example having seven primes, yielding the
family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently
56003, being the first member of this family, is the smallest prime with
this property.

Find the smallest prime which, by replacing part of the number (not
necessarily adjacent digits) with the same digit, is part of an eight
prime value family.


Problem 52
==========

It can be seen that the number, 125874, and its double, 251748, contain
exactly the same digits, but in a different order.

Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x,
contain the same digits.


Problem 53
==========

There are exactly ten ways of selecting three from five, 12345:

           123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, nCr(5,3) = 10.

In general,

nCr(n,r) = n!/(r!(n-r)!), where r =< n, n! = n * (n1) * ... * 3 * 2 * 1,
and 0! = 1.

It is not until n = 23, that a value exceeds one-million: nCr(23,10) =
1144066.

How many values of nCr(n,r), for 1 =< n =< 100, are greater than one-million?


Problem 54
==========

In the card game poker, a hand consists of five cards and are ranked, from
lowest to highest, in the following way:

  * High Card: Highest value card.
  * One Pair: Two cards of the same value.
  * Two Pairs: Two different pairs.
  * Three of a Kind: Three cards of the same value.
  * Straight: All cards are consecutive values.
  * Flush: All cards of the same suit.
  * Full House: Three of a kind and a pair.
  * Four of a Kind: Four cards of the same value.
  * Straight Flush: All cards are consecutive values of same suit.
  * Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.

The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.

If two players have the same ranked hands then the rank made up of the
highest value wins; for example, a pair of eights beats a pair of fives
(see example 1 below). But if two ranks tie, for example, both players
have a pair of queens, then highest cards in each hand are compared (see
example 4 below); if the highest cards tie then the next highest cards are
compared, and so on.

Consider the following five hands dealt to two players:

        Hand   Player 1            Player 2              Winner
        1      5H 5C 6S 7S KD      2C 3S 8S 8D TD        Player 2
               Pair of Fives       Pair of Eights
        2      5D 8C 9S JS AC      2C 5C 7D 8S QH        Player 1
               Highest card Ace    Highest card Queen
        3      2D 9C AS AH AC      3D 6D 7D TD QD        Player 2
               Three Aces          Flush with Diamonds
               4D 6S 9H QH QC      3D 6D 7H QD QS
        4      Pair of Queens      Pair of Queens        Player 1
               Highest card Nine   Highest card Seven
               2H 2D 4C 4D 4S      3C 3D 3S 9S 9D
        5      Full House          Full House            Player 1
               With Three Fours    with Three Threes

The file poker.txt contains one-thousand random hands dealt to two players.
Each line of the file contains ten cards (separated by a single space): the
first five are Player 1's cards and the last five are Player 2's cards. You
can assume that all hands are valid (no invalid characters or repeated
cards), each player's hand is in no specific order, and in each hand there
is a clear winner.

How many hands does Player 1 win?


Problem 55
==========

If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.

Not all numbers produce palindromes so quickly. For example,

349 + 943 = 1292,
1292 + 2921 = 4213
4213 + 3124 = 7337

That is, 349 took three iterations to arrive at a palindrome.

Although no one has proved it yet, it is thought that some numbers, like
196, never produce a palindrome. A number that never forms a palindrome
through the reverse and add process is called a Lychrel number. Due to the
theoretical nature of these numbers, and for the purpose of this problem,
we shall assume that a number is Lychrel until proven otherwise. In
addition you are given that for every number below ten-thousand, it will
either (i) become a palindrome in less than fifty iterations, or, (ii) no
one, with all the computing power that exists, has managed so far to map
it to a palindrome. In fact, 10677 is the first number to be shown to
require over fifty iterations before producing a palindrome:
4668731596684224866951378664 (53 iterations, 28-digits).

Surprisingly, there are palindromic numbers that are themselves Lychrel
numbers; the first example is 4994.

How many Lychrel numbers are there below ten-thousand?

NOTE: Wording was modified slightly on 24 April 2007 to emphasise the
theoretical nature of Lychrel numbers.


Problem 56
==========

A googol (10^100) is a massive number: one followed by one-hundred zeros;
100^100 is almost unimaginably large: one followed by two-hundred zeros.
Despite their size, the sum of the digits in each number is only 1.

Considering natural numbers of the form, a^b, where a, b < 100, what is
the maximum digital sum?


Problem 57
==========

It is possible to show that the square root of two can be expressed as an
infinite continued fraction.

            2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...

The next three expansions are 99/70, 239/169, and 577/408, but the eighth
expansion, 1393/985, is the first example where the number of digits in
the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a
numerator with more digits than denominator?


Problem 58
==========

Starting with 1 and spiralling anticlockwise in the following way, a
square spiral with side length 7 is formed.

                           37 36 35 34 33 32 31
                           38 17 16 15 14 13 30
                           39 18  5  4  3 12 29
                           40 19  6  1  2 11 28
                           41 20  7  8  9 10 27
                           42 21 22 23 24 25 26
                           43 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right
diagonal, but what is more interesting is that 8 out of the 13 numbers
lying along both diagonals are prime; that is, a ratio of 8/13 62%.

If one complete new layer is wrapped around the spiral above, a square
spiral with side length 9 will be formed. If this process is continued,
what is the side length of the square spiral for which the ratio of primes
along both diagonals first falls below 10%?


Problem 59
==========

Each character on a computer is assigned a unique code and the preferred
standard is ASCII (American Standard Code for Information Interchange).
For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.

A modern encryption method is to take a text file, convert the bytes to
ASCII, then XOR each byte with a given value, taken from a secret key. The
advantage with the XOR function is that using the same encryption key on
the cipher text, restores the plain text; for example, 65 XOR 42 = 107,
then 107 XOR 42 = 65.

For unbreakable encryption, the key is the same length as the plain text
message, and the key is made up of random bytes. The user would keep the
encrypted message and the encryption key in different locations, and
without both "halves", it is impossible to decrypt the message.

Unfortunately, this method is impractical for most users, so the modified
method is to use a password as a key. If the password is shorter than the
message, which is likely, the key is repeated cyclically throughout the
message. The balance for this method is using a sufficiently long password
key for security, but short enough to be memorable.

Your task has been made easy, as the encryption key consists of three
lower case characters. Using cipher1.txt, a file containing the encrypted
ASCII codes, and the knowledge that the plain text must contain common
English words, decrypt the message and find the sum of the ASCII values
in the original text.


Problem 60
==========

The primes 3, 7, 109, and 673, are quite remarkable. By taking any two
primes and concatenating them in any order the result will always be
prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The
sum of these four primes, 792, represents the lowest sum for a set of four
primes with this property.

Find the lowest sum for a set of five primes for which any two primes
concatenate to produce another prime.


Problem 61
==========

Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers
are all figurate (polygonal) numbers and are generated by the following
formulae:

Triangle     P[3,n]=n(n+1)/2    1, 3, 6, 10, 15, ...
Square       P[4,n]=n^2         1, 4, 9, 16, 25, ...
Pentagonal   P[5,n]=n(3n-1)/2   1, 5, 12, 22, 35, ...
Hexagonal    P[6,n]=n(2n-1)     1, 6, 15, 28, 45, ...
Heptagonal   P[7,n]=n(5n-3)/2   1, 7, 18, 34, 55, ...
Octagonal    P[8,n]=n(3n-2)     1, 8, 21, 40, 65, ...

The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three
interesting properties.

 1. The set is cyclic, in that the last two digits of each number is the
    first two digits of the next number (including the last number with
    the first).
 2. Each polygonal type: triangle (P[3,127]=8128), square (P[4,91]=8281),
    and pentagonal (P[5,44]=2882), is represented by a different number in
    the set.
 3. This is the only set of 4-digit numbers with this property.

Find the sum of the only ordered set of six cyclic 4-digit numbers for
which each polygonal type: triangle, square, pentagonal, hexagonal,
heptagonal, and octagonal, is represented by a different number in the
set.


Problem 62
==========

The cube, 41063625 (345^3), can be permuted to produce two other cubes:
56623104 (384^3) and 66430125 (405^3). In fact, 41063625 is the smallest
cube which has exactly three permutations of its digits which are also
cube.

Find the smallest cube for which exactly five permutations of its digits
are cube.


Problem 63
==========

The 5-digit number, 16807=7^5, is also a fifth power. Similarly, the
9-digit number, 134217728=8^9, is a ninth power.

How many n-digit positive integers exist which are also an nth power?


Problem 64
==========

All square roots are periodic when written as continued fractions and can
be written in the form:

N = a[0] +            1
           a[1] +         1
                  a[2] +     1
                         a[3] + ...

For example, let us consider 23:

23 = 4 + 23 -- 4 = 4 +  1  = 4 +  1     1     1 +  23 - 3
                                      23--4          7

If we continue we would get the following expansion:

23 = 4 +          1
         1 +        1
             3 +      1
                 1 +    1
                     8 + ...

The process can be summarised as follows:

a[0] = 4,     1    =   23+4    = 1 +  23--3
            23--4        7              7
a[1] = 1,     7    =  7(23+3)  = 3 +  23--3
            23--3       14              2
a[2] = 3,     2    =  2(23+3)  = 1 +  23--4
            23--3       14              7
a[3] = 1,     7    =  7(23+4)  = 8 +  23--4
            23--4        7
a[4] = 8,     1    =   23+4    = 1 +  23--3
            23--4        7              7
a[5] = 1,     7    =  7(23+3)  = 3 +  23--3
            23--3       14              2
a[6] = 3,     2    =  2(23+3)  = 1 +  23--4
            23--3       14              7
a[7] = 1,     7    =  7(23+4)  = 8 +  23--4
            23--4        7

It can be seen that the sequence is repeating. For conciseness, we use the
notation 23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats
indefinitely.

The first ten continued fraction representations of (irrational) square
roots are:

2=[1;(2)], period=1
3=[1;(1,2)], period=2
5=[2;(4)], period=1
6=[2;(2,4)], period=2
7=[2;(1,1,1,4)], period=4
8=[2;(1,4)], period=2
10=[3;(6)], period=1
11=[3;(3,6)], period=2
12= [3;(2,6)], period=2
13=[3;(1,1,1,1,6)], period=5

Exactly four continued fractions, for N 13, have an odd period.

How many continued fractions for N 10000 have an odd period?


Problem 65
==========

The square root of 2 can be written as an infinite continued fraction.

2 = 1 +          1
        2 +        1
            2 +      1
                2 +    1
                    2 + ...

The infinite continued fraction can be written, 2 = [1;(2)], (2) indicates
that 2 repeats ad infinitum. In a similar way, 23 = [4;(1,3,1,8)].

It turns out that the sequence of partial values of continued fractions
for square roots provide the best rational approximations. Let us consider
the convergents for 2.

1 + 1 = 3/2
    2

1 +   1   = 7/5
    2 + 1
        2

1 +     1     = 17/12
    2 +   1
        2 + 1
            2

1 +       1       = 41/29
    2 +     1
        2 +   1
            2 + 1
                2

Hence the sequence of the first ten convergents for 2 are:
1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378,
...

What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].

The first ten terms in the sequence of convergents for e are:
2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...

The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100th convergent of the
continued fraction for e.


Problem 66
==========

Consider quadratic Diophantine equations of the form:

                              x^2 - Dy^2 = 1

For example, when D=13, the minimal solution in x is 649^2 - 13 * 180^2 =
1.

It can be assumed that there are no solutions in positive integers when D
is square.

By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the
following:

3^2 - 2 * 2^2 = 1
2^2 - 3 * 1^2 = 1
9^2 - 5 * 4^2 = 1
5^2 - 6 * 2^2 = 1
8^2 - 7 * 3^2 = 1

Hence, by considering minimal solutions in x for D 7, the largest x is
obtained when D=5.

Find the value of D 1000 in minimal solutions of x for which the largest
value of x is obtained.


Problem 67
==========

By starting at the top of the triangle below and moving to adjacent
numbers on the row below, the maximum total from top to bottom is 23.

                                    3
                                   7 4
                                  2 4 6
                                 8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom in triangle.txt, a 15K text file
containing a triangle with one-hundred rows.

NOTE: This is a much more difficult version of Problem 18. It is not
possible to try every route to solve this problem, as there are 2^99
altogether! If you could check one trillion (10^12) routes every second it
would take over twenty billion years to check them all. There is an
efficient algorithm to solve it. ;o)


Problem 68
==========

Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6,
and each line adding to nine.

Working clockwise, and starting from the group of three with the
numerically lowest external node (4,3,2 in this example), each solution
can be described uniquely. For example, the above solution can be
described by the set: 4,3,2; 6,2,1; 5,1,3.

It is possible to complete the ring with four different totals: 9, 10, 11,
and 12. There are eight solutions in total.

        Total          Solution Set
        9              4,2,3; 5,3,1; 6,1,2
        9              4,3,2; 6,2,1; 5,1,3
        10             2,3,5; 4,5,1; 6,1,3
        10             2,5,3; 6,3,1; 4,1,5
        11             1,4,6; 3,6,2; 5,2,4
        11             1,6,4; 5,4,2; 3,2,6
        12             1,5,6; 2,6,4; 3,4,5
        12             1,6,5; 3,5,4; 2,4,6

By concatenating each group it is possible to form 9-digit strings; the
maximum string for a 3-gon ring is 432621513.

Using the numbers 1 to 10, and depending on arrangements, it is possible
to form 16- and 17-digit strings. What is the maximum 16-digit string for
a "magic" 5-gon ring?


Problem 69
==========

Euler's Totient function, f(n) [sometimes called the phi function], is
used to determine the number of numbers less than n which are relatively
prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine
and relatively prime to nine, f(9)=6.

+------------------------------------------+
| n  | Relatively Prime | f(n) | n/f(n)    |
|----+------------------+------+-----------|
| 2  | 1                | 1    | 2         |
|----+------------------+------+-----------|
| 3  | 1,2              | 2    | 1.5       |
|----+------------------+------+-----------|
| 4  | 1,3              | 2    | 2         |
|----+------------------+------+-----------|
| 5  | 1,2,3,4          | 4    | 1.25      |
|----+------------------+------+-----------|
| 6  | 1,5              | 2    | 3         |
|----+------------------+------+-----------|
| 7  | 1,2,3,4,5,6      | 6    | 1.1666... |
|----+------------------+------+-----------|
| 8  | 1,3,5,7          | 4    | 2         |
|----+------------------+------+-----------|
| 9  | 1,2,4,5,7,8      | 6    | 1.5       |
|----+------------------+------+-----------|
| 10 | 1,3,7,9          | 4    | 2.5       |
+------------------------------------------+

It can be seen that n=6 produces a maximum n/f(n) for n 10.

Find the value of n 1,000,000 for which n/f(n) is a maximum.


Problem 70
==========

Euler's Totient function, f(n) [sometimes called the phi function], is
used to determine the number of positive numbers less than or equal to n
which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are
all less than nine and relatively prime to nine, f(9)=6.
The number 1 is considered to be relatively prime to every positive
number, so f(1)=1.

Interestingly, f(87109)=79180, and it can be seen that 87109 is a
permutation of 79180.

Find the value of n, 1 < n < 10^7, for which f(n) is a permutation of n
and the ratio n/f(n) produces a minimum.


Problem 71
==========

Consider the fraction, n/d, where n and d are positive integers. If n < d
and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d 8 in ascending order
of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3,
                       5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that 2/5 is the fraction immediately to the left of 3/7.

By listing the set of reduced proper fractions for d 1,000,000 in
ascending order of size, find the numerator of the fraction immediately to
the left of 3/7.


Problem 72
==========

Consider the fraction, n/d, where n and d are positive integers. If n < d
and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d 8 in ascending order
of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3,
                       5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that there are 21 elements in this set.

How many elements would be contained in the set of reduced proper
fractions for d 1,000,000?


Problem 73
==========

Consider the fraction, n/d, where n and d are positive integers. If n < d
and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d 8 in ascending order
of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3,
                       5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that there are 3 fractions between 1/3 and 1/2.

How many fractions lie between 1/3 and 1/2 in the sorted set of reduced
proper fractions for d 10,000?


Problem 74
==========

The number 145 is well known for the property that the sum of the
factorial of its digits is equal to 145:

1! + 4! + 5! = 1 + 24 + 120 = 145

Perhaps less well known is 169, in that it produces the longest chain of
numbers that link back to 169; it turns out that there are only three such
loops that exist:

169 363601 1454 169
871 45361 871
872 45362 872

It is not difficult to prove that EVERY starting number will eventually
get stuck in a loop. For example,

69 363600 1454 169 363601 ( 1454)
78 45360 871 45361 ( 871)
540 145 ( 145)

Starting with 69 produces a chain of five non-repeating terms, but the
longest non-repeating chain with a starting number below one million is
sixty terms.

How many chains, with a starting number below one million, contain exactly
sixty non-repeating terms?


Problem 75
==========

It turns out that 12 cm is the smallest length of wire can be bent to form
a right angle triangle in exactly one way, but there are many more
examples.

12 cm: (3,4,5)
24 cm: (6,8,10)
30 cm: (5,12,13)
36 cm: (9,12,15)
40 cm: (8,15,17)
48 cm: (12,16,20)

In contrast, some lengths of wire, like 20 cm, cannot be bent to form a
right angle triangle, and other lengths allow more than one solution to be
found; for example, using 120 cm it is possible to form exactly three
different right angle triangles.

120 cm: (30,40,50), (20,48,52), (24,45,51)

Given that L is the length of the wire, for how many values of L 2,000,000
can exactly one right angle triangle be formed?


Problem 76
==========

It is possible to write five as a sum in exactly six different ways:

4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1

How many different ways can one hundred be written as a sum of at least
two positive integers?


Problem 77
==========

It is possible to write ten as the sum of primes in exactly five different
ways:

7 + 3
5 + 5
5 + 3 + 2
3 + 3 + 2 + 2
2 + 2 + 2 + 2 + 2

What is the first value which can be written as the sum of primes in over
five thousand different ways?


Problem 78
==========

Let p(n) represent the number of different ways in which n coins can be
separated into piles. For example, five coins can separated into piles in
exactly seven different ways, so p(5)=7.

                            OOOOO

                            OOOO   O

                            OOO   OO

                            OOO   O   O

                            OO   OO   O

                            OO   O   O   O

                            O   O   O   O   O

Find the least value of n for which p(n) is divisible by one million.


Problem 79
==========

A common security method used for online banking is to ask the user for
three random characters from a passcode. For example, if the passcode was
531278, they may asked for the 2nd, 3rd, and 5th characters; the expected
reply would be: 317.

The text file keylog.txt contains fifty successful login attempts.

Given that the three characters are always asked for in order, analyse the
file so as to determine the shortest possible secret passcode of unknown
length.


Problem 80
==========

It is well known that if the square root of a natural number is not an
integer, then it is irrational. The decimal expansion of such square roots
is infinite without any repeating pattern at all.

The square root of two is 1.41421356237309504880..., and the digital sum
of the first one hundred decimal digits is 475.

For the first one hundred natural numbers, find the total of the digital
sums of the first one hundred decimal digits for all the irrational square
roots.


Problem 81
==========

In the 5 by 5 matrix below, the minimal path sum from the top left to the
bottom right, by only moving to the right and down, is indicated in red
and is equal to 2427.

                           131 673 234 103 18
                           201 96  342 965 150
                           630 803 746 422 111
                           537 699 497 121 956
                           805 732 524 37  331

Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by
80 matrix, from the top left to the bottom right by only moving right and down.


Problem 82
==========

   NOTE: This problem is a more challenging version of Problem 81.

The minimal path sum in the 5 by 5 matrix below, by starting in any cell
in the left column and finishing in any cell in the right column, and only
moving up, down, and right, is indicated in red; the sum is equal to 994.

                           131 673 234 103 18
                           201 96  342 965 150
                           630 803 746 422 111
                           537 699 497 121 956
                           805 732 524 37  331

Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by
80 matrix, from the left column to the right column.


Problem 83
==========

NOTE: This problem is a significantly more challenging version of
   Problem 81.

In the 5 by 5 matrix below, the minimal path sum from the top left to the
bottom right, by moving left, right, up, and down, is indicated in red and
is equal to 2297.

                           131 673 234 103 18
                           201 96  342 965 150
                           630 803 746 422 111
                           537 699 497 121 956
                           805 732 524 37  331

Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by
80 matrix, from the top left to the bottom right by moving left, right, up,
and down.


Problem 84
==========

In the game, Monopoly, the standard board is set up in the following way:

             GO   A1  CC1  A2  T1  R1  B1  CH1  B2   B3  JAIL
             H2                                          C1
             T2                                          U1
             H1                                          C2
             CH3                                         C3
             R4                                          R2
             G3                                          D1
             CC3                                         CC2
             G2                                          D2
             G1                                          D3
             G2J  F3  U2   F2  F1  R3  E3  E2   CH2  E1  FP

A player starts on the GO square and adds the scores on two 6-sided dice
to determine the number of squares they advance in a clockwise direction.
Without any further rules we would expect to visit each square with equal
probability: 2.5%. However, landing on G2J (Go To Jail), CC (community
chest), and CH (chance) changes this distribution.

In addition to G2J, and one card from each of CC and CH, that orders the
player to go to directly jail, if a player rolls three consecutive
doubles, they do not advance the result of their 3rd roll. Instead they
proceed directly to jail.

At the beginning of the game, the CC and CH cards are shuffled. When a
player lands on CC or CH they take a card from the top of the respective
pile and, after following the instructions, it is returned to the bottom
of the pile. There are sixteen cards in each pile, but for the purpose of
this problem we are only concerned with cards that order a movement; any
instruction not concerned with movement will be ignored and the player
will remain on the CC/CH square.

  * Community Chest (2/16 cards):

      1. Advance to GO
      2. Go to JAIL

  * Chance (10/16 cards):

      1. Advance to GO
      2. Go to JAIL
      3. Go to C1
      4. Go to E3
      5. Go to H2
      6. Go to R1
      7. Go to next R (railway company)
      8. Go to next R
      9. Go to next U (utility company)
     10. Go back 3 squares.

The heart of this problem concerns the likelihood of visiting a particular
square. That is, the probability of finishing at that square after a roll.
For this reason it should be clear that, with the exception of G2J for
which the probability of finishing on it is zero, the CH squares will have
the lowest probabilities, as 5/8 request a movement to another square, and
it is the final square that the player finishes at on each roll that we
are interested in. We shall make no distinction between "Just Visiting"
and being sent to JAIL, and we shall also ignore the rule about requiring
a double to "get out of jail", assuming that they pay to get out on their
next turn.

By starting at GO and numbering the squares sequentially from 00 to 39 we
can concatenate these two-digit numbers to produce strings that correspond
with sets of squares.

Statistically it can be shown that the three most popular squares, in
order, are JAIL (6.24%) = Square 10, E3 (3.18%) = Square 24, and GO
(3.09%) = Square 00. So these three most popular squares can be listed
with the six-digit modal string: 102400.

If, instead of using two 6-sided dice, two 4-sided dice are used, find the
six-digit modal string.


Problem 85
==========

By counting carefully it can be seen that a rectangular grid measuring 3
by 2 contains eighteen rectangles:

Although there exists no rectangular grid that contains exactly two
million rectangles, find the area of the grid with the nearest solution.


Problem 86
==========

A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3,
and a fly, F, sits in the opposite corner. By travelling on the surfaces
of the room the shortest "straight line" distance from S to F is 10 and
the path is shown on the diagram.

However, there are up to three "shortest" path candidates for any given
cuboid and the shortest route is not always integer.

By considering all cuboid rooms up to a maximum size of M by M by M, there
are exactly 2060 cuboids for which the shortest distance is integer when
M=100, and this is the least value of M for which the number of solutions
first exceeds two thousand; the number of solutions is 1975 when M=99.

Find the least value of M such that the number of solutions first exceeds
one million.


Problem 87
==========

The smallest number expressible as the sum of a prime square, prime cube,
and prime fourth power is 28. In fact, there are exactly four numbers
below fifty that can be expressed in such a way:

28 = 2^2 + 2^3 + 2^4
33 = 3^2 + 2^3 + 2^4
49 = 5^2 + 2^3 + 2^4
47 = 2^2 + 3^3 + 2^4

How many numbers below fifty million can be expressed as the sum of a
prime square, prime cube, and prime fourth power?


Problem 88
==========

A natural number, N, that can be written as the sum and product of a given
set of at least two natural numbers, {a[1], a[2], ... , a[k]} is called a
product-sum number: N = a[1] + a[2] + ... + a[k] = a[1] * a[2] * ... *
a[k].

For example, 6 = 1 + 2 + 3 = 1 * 2 * 3.

For a given set of size, k, we shall call the smallest N with this
property a minimal product-sum number. The minimal product-sum numbers for
sets of size, k = 2, 3, 4, 5, and 6 are as follows.

k=2: 4 = 2 * 2 = 2 + 2
k=3: 6 = 1 * 2 * 3 = 1 + 2 + 3
k=4: 8 = 1 * 1 * 2 * 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 * 1 * 2 * 2 * 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 * 1 * 1 * 1 * 2 * 6 = 1 + 1 + 1 + 1 + 2 + 6

Hence for 2k6, the sum of all the minimal product-sum numbers is 4+6+8+12
= 30; note that 8 is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2k12 is
{4, 6, 8, 12, 15, 16}, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2k12000?


Problem 89
==========

The rules for writing Roman numerals allow for many ways of writing each
number. However, there is always a "best" way of writing a particular number.

For example, the following represent all of the legitimate ways of writing
the number sixteen:

IIIIIIIIIIIIIIII
VIIIIIIIIIII
VVIIIIII
XIIIIII
VVVI
XVI

The last example being considered the most efficient, as it uses the least
number of numerals.

The 11K text file roman.txt contains one thousand numbers written in valid,
but not necessarily minimal, Roman numerals; that is, they are arranged in
descending units and obey the subtractive pair rule (see FAQ for the
definitive rules for this problem).

Find the number of characters saved by writing each of these in their
minimal form.

Note: You can assume that all the Roman numerals in the file contain no
more than four consecutive identical units.

FAQ Link: http://projecteuler.net/about=roman_numerals


Problem 90
==========

Each of the six faces on a cube has a different digit (0 to 9) written on
it; the same is done to a second cube. By placing the two cubes
side-by-side in different positions we can form a variety of 2-digit
numbers.

For example, the square number 64 could be formed:

In fact, by carefully choosing the digits on both cubes it is possible to
display all of the square numbers below one-hundred: 01, 04, 09, 16, 25,
36, 49, 64, and 81.

For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9}
on one cube and {1, 2, 3, 4, 8, 9} on the other cube.

However, for this problem we shall allow the 6 or 9 to be turned
upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3,
4, 6, 7} allows for all nine square numbers to be displayed; otherwise it
would be impossible to obtain 09.

In determining a distinct arrangement we are interested in the digits on
each cube, not the order.

{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}
{1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}

But because we are allowing 6 and 9 to be reversed, the two distinct sets
in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9}
for the purpose of forming 2-digit numbers.

How many distinct arrangements of the two cubes allow for all of the
square numbers to be displayed?


Problem 91
==========

The points P (x[1], y[1]) and Q (x[2], y[2]) are plotted at integer
co-ordinates and are joined to the origin, O(0,0), to form DOPQ.

There are exactly fourteen triangles containing a right angle that can be
formed when each co-ordinate lies between 0 and 2 inclusive; that is,
0 x[1], y[1], x[2], y[2] 2.

Given that 0 x[1], y[1], x[2], y[2] 50, how many right triangles can be
formed?


Problem 92
==========

A number chain is created by continuously adding the square of the digits
in a number to form a new number until it has been seen before.

For example,

44 32 13 10 1 1
85 89 145 42 20 4 16 37 58 89

Therefore any chain that arrives at 1 or 89 will become stuck in an
endless loop. What is most amazing is that EVERY starting number will
eventually arrive at 1 or 89.

How many starting numbers below ten million will arrive at 89?


Problem 93
==========

By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and
making use of the four arithmetic operations (+, , *, /) and
brackets/parentheses, it is possible to form different positive integer
targets.

For example,

8 = (4 * (1 + 3)) / 2
14 = 4 * (3 + 1 / 2)
19 = 4 * (2 + 3) 1
36 = 3 * 4 * (2 + 1)

Note that concatenations of the digits, like 12 + 34, are not allowed.

Using the set, {1, 2, 3, 4}, it is possible to obtain thirty-one different
target numbers of which 36 is the maximum, and each of the numbers 1 to 28
can be obtained before encountering the first non-expressible number.

Find the set of four distinct digits, a < b < c < d, for which the longest
set of consecutive positive integers, 1 to n, can be obtained, giving your
answer as a string: abcd.


Problem 94
==========

It is easily proved that no equilateral triangle exists with integral
length sides and integral area. However, the almost equilateral triangle
5-5-6 has an area of 12 square units.

We shall define an almost equilateral triangle to be a triangle for which
two sides are equal and the third differs by no more than one unit.

Find the sum of the perimeters of every almost equilateral triangle with
integral side lengths and area and whose perimeters do not exceed one
billion (1,000,000,000).


Problem 95
==========

The proper divisors of a number are all the divisors excluding the number
itself. For example, the proper divisors of 28 are 1, 2, 4, 7, and 14. As
the sum of these divisors is equal to 28, we call it a perfect number.

Interestingly the sum of the proper divisors of 220 is 284 and the sum of
the proper divisors of 284 is 220, forming a chain of two numbers. For
this reason, 220 and 284 are called an amicable pair.

Perhaps less well known are longer chains. For example, starting with
12496, we form an amicable chain of five numbers:

                12496 14288 15472 14536 14264 ( 12496 ...)

Find the smallest member of the longest amicable chain with no element
exceeding one million.


Problem 96
==========

Su Doku (Japanese meaning number place) is the name given to a popular
puzzle concept. Its origin is unclear, but credit must be attributed to
Leonhard Euler who invented a similar, and much more difficult, puzzle
idea called Latin Squares. The objective of Su Doku puzzles, however, is
to replace the blanks (or zeros) in a 9 by 9 grid in such that each row,
column, and 3 by 3 box contains each of the digits 1 to 9. Below is an
example of a typical starting puzzle grid and its solution grid.

       +-----------------------+         +-----------------------+
       | 0 0 3 | 0 2 0 | 6 0 0 |         | 4 8 3 | 9 2 1 | 6 5 7 |
       | 9 0 0 | 3 0 5 | 0 0 1 |         | 9 6 7 | 3 4 5 | 8 2 1 |
       | 0 0 1 | 8 0 6 | 4 0 0 |         | 2 5 1 | 8 7 6 | 4 9 3 |
       |-------+-------+-------|         |-------+-------+-------|
       | 0 0 8 | 1 0 2 | 9 0 0 |         | 5 4 8 | 1 3 2 | 9 7 6 |
       | 7 0 0 | 0 0 0 | 0 0 8 |         | 7 2 9 | 5 6 4 | 1 3 8 |
       | 0 0 6 | 7 0 8 | 2 0 0 |         | 1 3 6 | 7 9 8 | 2 4 5 |
       |-------+-------+-------|         |-------+-------+-------|
       | 0 0 2 | 6 0 9 | 5 0 0 |         | 3 7 2 | 6 8 9 | 5 1 4 |
       | 8 0 0 | 2 0 3 | 0 0 9 |         | 8 1 4 | 2 5 3 | 7 6 9 |
       | 0 0 5 | 0 1 0 | 3 0 0 |         | 6 9 5 | 4 1 7 | 3 8 2 |
       +-----------------------+         +-----------------------+

A well constructed Su Doku puzzle has a unique solution and can be solved
by logic, although it may be necessary to employ "guess and test" methods
in order to eliminate options (there is much contested opinion over this).
The complexity of the search determines the difficulty of the puzzle; the
example above is considered easy because it can be solved by straight
forward direct deduction.

The 6K text file sudoku.txt contains fifty different Su Doku puzzles ranging
in difficulty, but all with unique solutions (the first puzzle in the file is
the example above).

By solving all fifty puzzles find the sum of the 3-digit numbers found in
the top left corner of each solution grid; for example, 483 is the 3-digit
number found in the top left corner of the solution grid above.


Problem 97
==========

The first known prime found to exceed one million digits was discovered in
1999, and is a Mersenne prime of the form 2^69725931; it contains exactly
2,098,960 digits. Subsequently other Mersenne primes, of the form 2^p1,
have been found which contain more digits.

However, in 2004 there was found a massive non-Mersenne prime which
contains 2,357,207 digits: 28433 * 2^7830457+1.

Find the last ten digits of this prime number.


Problem 98
==========

By replacing each of the letters in the word CARE with 1, 2, 9, and 6
respectively, we form a square number: 1296 = 36^2. What is remarkable is
that, by using the same digital substitutions, the anagram, RACE, also
forms a square number: 9216 = 96^2. We shall call CARE (and RACE) a square
anagram word pair and specify further that leading zeroes are not
permitted, neither may a different letter have the same digital value as
another letter.

Using words.txt, a 16K text file containing nearly two-thousand common English
words, find all the square anagram word pairs (a palindromic word is NOT
considered to be an anagram of itself).

What is the largest square number formed by any member of such a pair?

NOTE: All anagrams formed must be contained in the given text file.


Problem 99
==========

Comparing two numbers written in index form like 2^11 and 3^7 is not
difficult, as any calculator would confirm that 2^11 = 2048 < 3^7 = 2187.

However, confirming that 632382^518061 > 519432^525806 would be much more
difficult, as both numbers contain over three million digits.

Using base_exp.txt, a 22K text file containing one thousand lines with a
base/exponent pair on each line, determine which line number has the
greatest numerical value.

NOTE: The first two lines in the file represent the numbers in the example
given above.


Problem 100
===========

If a box contains twenty-one coloured discs, composed of fifteen blue
discs and six red discs, and two discs were taken at random, it can be
seen that the probability of taking two blue discs, P(BB) = (15/21) *
(14/20) = 1/2.

The next such arrangement, for which there is exactly 50% chance of taking
two blue discs at random, is a box containing eighty-five blue discs and
thirty-five red discs.

By finding the first arrangement to contain over 10^12 = 1,000,000,000,000
discs in total, determine the number of blue discs that the box would
contain.


Problem 101
===========

If we are presented with the first k terms of a sequence it is impossible
to say with certainty the value of the next term, as there are infinitely
many polynomial functions that can model the sequence.

As an example, let us consider the sequence of cube numbers. This is
defined by the generating function, u[n] = n^3: 1, 8, 27, 64, 125, 216,
...

Suppose we were only given the first two terms of this sequence. Working
on the principle that "simple is best" we should assume a linear
relationship and predict the next term to be 15 (common difference 7).
Even if we were presented with the first three terms, by the same
principle of simplicity, a quadratic relationship should be assumed.

We shall define OP(k, n) to be the nth term of the optimum polynomial
generating function for the first k terms of a sequence. It should be
clear that OP(k, n) will accurately generate the terms of the sequence for
n k, and potentially the first incorrect term (FIT) will be OP(k, k+1); in
which case we shall call it a bad OP (BOP).

As a basis, if we were only given the first term of sequence, it would be
most sensible to assume constancy; that is, for n 2, OP(1, n) = u[1].

Hence we obtain the following OPs for the cubic sequence:

OP(1, n) = 1              1, 1, 1, 1, ...
OP(2, n) = 7n6            1, 8, 15, ...
OP(3, n) = 6n^211n+6      1, 8, 27, 58, ...
OP(4, n) = n^3            1, 8, 27, 64, 125, ...

Clearly no BOPs exist for k 4.

By considering the sum of FITs generated by the BOPs (indicated in red
above), we obtain 1 + 15 + 58 = 74.

Consider the following tenth degree polynomial generating function:

        u[n] = 1 n + n^2 n^3 + n^4 n^5 + n^6 n^7 + n^8 n^9 + n^10

Find the sum of FITs for the BOPs.


Problem 102
===========

Three distinct points are plotted at random on a Cartesian plane, for
which -1000 x, y 1000, such that a triangle is formed.

Consider the following two triangles:

                  A(-340,495), B(-153,-910), C(835,-947)

                  X(-175,41), Y(-421,-714), Z(574,-645)

It can be verified that triangle ABC contains the origin, whereas triangle
XYZ does not.

Using triangles.txt, a 27K text file containing the co-ordinates of one
thousand "random" triangles, find the number of triangles for which the
interior contains the origin.

NOTE: The first two examples in the file represent the triangles in the
example given above.


Problem 103
===========

Let S(A) represent the sum of elements in set A of size n. We shall call
it a special sum set if for any two non-empty disjoint subsets, B and C,
the following properties are true:

 1. S(B) S(C); that is, sums of subsets cannot be equal.
 2. If B contains more elements than C then S(B) > S(C).

If S(A) is minimised for a given n, we shall call it an optimum special
sum set. The first five optimum special sum sets are given below.

n = 1: {1}
n = 2: {1, 2}
n = 3: {2, 3, 4}
n = 4: {3, 5, 6, 7}
n = 5: {6, 9, 11, 12, 13}

It seems that for a given optimum set, A = {a[1], a[2], ... , a[n]}, the
next optimum set is of the form B = {b, a[1]+b, a[2]+b, ... ,a[n]+b},
where b is the "middle" element on the previous row.

By applying this "rule" we would expect the optimum set for n = 6 to be A
= {11, 17, 20, 22, 23, 24}, with S(A) = 117. However, this is not the
optimum set, as we have merely applied an algorithm to provide a near
optimum set. The optimum set for n = 6 is A = {11, 18, 19, 20, 22, 25},
with S(A) = 115 and corresponding set string: 111819202225.

Given that A is an optimum special sum set for n = 7, find its set string.

NOTE: This problem is related to problems 105 and 106.


Problem 104
===========

The Fibonacci sequence is defined by the recurrence relation:

  F[n] = F[n[1]] + F[n[2]], where F[1] = 1 and F[2] = 1.

It turns out that F[541], which contains 113 digits, is the first
Fibonacci number for which the last nine digits are 1-9 pandigital
(contain all the digits 1 to 9, but not necessarily in order). And
F[2749], which contains 575 digits, is the first Fibonacci number for
which the first nine digits are 1-9 pandigital.

Given that F[k] is the first Fibonacci number for which the first nine
digits AND the last nine digits are 1-9 pandigital, find k.


Problem 105
===========

Let S(A) represent the sum of elements in set A of size n. We shall call
it a special sum set if for any two non-empty disjoint subsets, B and C,
the following properties are true:

 1. S(B) S(C); that is, sums of subsets cannot be equal.
 2. If B contains more elements than C then S(B) > S(C).

For example, {81, 88, 75, 42, 87, 84, 86, 65} is not a special sum set
because 65 + 87 + 88 = 75 + 81 + 84, whereas {157, 150, 164, 119, 79, 159,
161, 139, 158} satisfies both rules for all possible subset pair
combinations and S(A) = 1286.

Using sets.txt, a 4K text file with one-hundred sets containing seven to
twelve elements (the two examples given above are the first two sets in the
file), identify all the special sum sets, A[1], A[2], ..., A[k], and find the
value of S(A[1]) + S(A[2]) + ... + S(A[k]).

NOTE: This problem is related to problems 103 and 106.


Problem 106
===========

Let S(A) represent the sum of elements in set A of size n. We shall call
it a special sum set if for any two non-empty disjoint subsets, B and C,
the following properties are true:

 1. S(B) S(C); that is, sums of subsets cannot be equal.
 2. If B contains more elements than C then S(B) > S(C).

For this problem we shall assume that a given set contains n strictly
increasing elements and it already satisfies the second rule.

Surprisingly, out of the 25 possible subset pairs that can be obtained
from a set for which n = 4, only 1 of these pairs need to be tested for
equality (first rule). Similarly, when n = 7, only 70 out of the 966
subset pairs need to be tested.

For n = 12, how many of the 261625 subset pairs that can be obtained need
to be tested for equality?

NOTE: This problem is related to problems 103 and 105.


Problem 107
===========

The following undirected network consists of seven vertices and twelve
edges with a total weight of 243.

The same network can be represented by the matrix below.

               +-----------------------------------------+
               |      | A  | B  | C  | D  | E  | F  | G  |
               |------+----+----+----+----+----+----+----|
               | A    | -  | 16 | 12 | 21 | -  | -  | -  |
               |------+----+----+----+----+----+----+----|
               | B    | 16 | -  | -  | 17 | 20 | -  | -  |
               |------+----+----+----+----+----+----+----|
               | C    | 12 | -  | -  | 28 | -  | 31 | -  |
               |------+----+----+----+----+----+----+----|
               | D    | 21 | 17 | 28 | -  | 18 | 19 | 23 |
               |------+----+----+----+----+----+----+----|
               | E    | -  | 20 | -  | 18 | -  | -  | 11 |
               |------+----+----+----+----+----+----+----|
               | F    | -  | -  | 31 | 19 | -  | -  | 27 |
               |------+----+----+----+----+----+----+----|
               | G    | -  | -  | -  | 23 | 11 | 27 | -  |
               +-----------------------------------------+

However, it is possible to optimise the network by removing some edges and
still ensure that all points on the network remain connected. The network
which achieves the maximum saving is shown below. It has a weight of 93,
representing a saving of 243 93 = 150 from the original network.

Using network.txt, a 6K text file containing a network with forty vertices,
and given in matrix form, find the maximum saving which can be achieved by
removing redundant edges whilst ensuring that the network remains connected.


Problem 108
===========

In the following equation x, y, and n are positive integers.

                                1 + 1 = 1
                                x   y   n

For n = 4 there are exactly three distinct solutions:

                                1 + 1  = 1
                                5   20   4
                                1 + 1  = 1
                                6   12   4
                                1 + 1  = 1
                                8   8    4

What is the least value of n for which the number of distinct solutions
exceeds one-thousand?

NOTE: This problem is an easier version of problem 110; it is strongly
advised that you solve this one first.


Problem 109
===========

In the game of darts a player throws three darts at a target board which
is split into twenty equal sized sections numbered one to twenty.

The score of a dart is determined by the number of the region that the
dart lands in. A dart landing outside the red/green outer ring scores
zero. The black and cream regions inside this ring represent single
scores. However, the red/green outer ring and middle ring score double and
treble scores respectively.

At the centre of the board are two concentric circles called the bull
region, or bulls-eye. The outer bull is worth 25 points and the inner bull
is a double, worth 50 points.

There are many variations of rules but in the most popular game the
players will begin with a score 301 or 501 and the first player to reduce
their running total to zero is a winner. However, it is normal to play a
"doubles out" system, which means that the player must land a double
(including the double bulls-eye at the centre of the board) on their final
dart to win; any other dart that would reduce their running total to one
or lower means the score for that set of three darts is "bust".

When a player is able to finish on their current score it is called a
"checkout" and the highest checkout is 170: T20 T20 D25 (two treble 20s
and double bull).

There are exactly eleven distinct ways to checkout on a score of 6:

                                +--------+
                                |D3|  |  |
                                |--+--+--|
                                |D1|D2|  |
                                |--+--+--|
                                |S2|D2|  |
                                |--+--+--|
                                |D2|D1|  |
                                |--+--+--|
                                |S4|D1|  |
                                |--+--+--|
                                |S1|S1|D2|
                                |--+--+--|
                                |S1|T1|D1|
                                |--+--+--|
                                |S1|S3|D1|
                                |--+--+--|
                                |D1|D1|D1|
                                |--+--+--|
                                |D1|S2|D1|
                                |--+--+--|
                                |S2|S2|D1|
                                +--------+

Note that D1 D2 is considered different to D2 D1 as they finish on
different doubles. However, the combination S1 T1 D1 is considered the
same as T1 S1 D1.

In addition we shall not include misses in considering combinations; for
example, D3 is the same as 0 D3 and 0 0 D3.

Incredibly there are 42336 distinct ways of checking out in total.

How many distinct ways can a player checkout with a score less than 100?


Problem 110
===========

In the following equation x, y, and n are positive integers.

                                1 + 1 = 1
                                x   y   n

It can be verified that when n = 1260 there are 113 distinct solutions and
this is the least value of n for which the total number of distinct
solutions exceeds one hundred.

What is the least value of n for which the number of distinct solutions
exceeds four million?

NOTE: This problem is a much more difficult version of problem 108 and
as it is well beyond the limitations of a brute force approach it requires
a clever implementation.


Problem 111
===========

Considering 4-digit primes containing repeated digits it is clear that
they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by
22, and so on. But there are nine 4-digit primes containing three ones:

           1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111

We shall say that M(n, d) represents the maximum number of repeated digits
for an n-digit prime where d is the repeated digit, N(n, d) represents the
number of such primes, and S(n, d) represents the sum of these primes.

So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit
prime where one is the repeated digit, there are N(4, 1) = 9 such primes,
and the sum of these primes is S(4, 1) = 22275. It turns out that for d =
0, it is only possible to have M(4, 0) = 2 repeated digits, but there are
N(4, 0) = 13 such cases.

In the same way we obtain the following results for 4-digit primes.

                +----------------------------------------+
                | Digit, d | M(4, d) | N(4, d) | S(4, d) |
                |----------+---------+---------+---------|
                | 0        | 2       | 13      | 67061   |
                |----------+---------+---------+---------|
                | 1        | 3       | 9       | 22275   |
                |----------+---------+---------+---------|
                | 2        | 3       | 1       | 2221    |
                |----------+---------+---------+---------|
                | 3        | 3       | 12      | 46214   |
                |----------+---------+---------+---------|
                | 4        | 3       | 2       | 8888    |
                |----------+---------+---------+---------|
                | 5        | 3       | 1       | 5557    |
                |----------+---------+---------+---------|
                | 6        | 3       | 1       | 6661    |
                |----------+---------+---------+---------|
                | 7        | 3       | 9       | 57863   |
                |----------+---------+---------+---------|
                | 8        | 3       | 1       | 8887    |
                |----------+---------+---------+---------|
                | 9        | 3       | 7       | 48073   |
                +----------------------------------------+

For d = 0 to 9, the sum of all S(4, d) is 273700.

Find the sum of all S(10, d).


Problem 112
===========

Working from left-to-right if no digit is exceeded by the digit to its
left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a
decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing
a "bouncy" number; for example, 155349.

Clearly there cannot be any bouncy numbers below one-hundred, but just
over half of the numbers below one-thousand (525) are bouncy. In fact, the
least number for which the proportion of bouncy numbers first reaches 50%
is 538.

Surprisingly, bouncy numbers become more and more common and by the time
we reach 21780 the proportion of bouncy numbers is equal to 90%.

Find the least number for which the proportion of bouncy numbers is
exactly 99%.


Problem 113
===========

Working from left-to-right if no digit is exceeded by the digit to its
left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a
decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing
a "bouncy" number; for example, 155349.

As n increases, the proportion of bouncy numbers below n increases such
that there are only 12951 numbers below one-million that are not bouncy
and only 277032 non-bouncy numbers below 10^10.

How many numbers below a googol (10^100) are not bouncy?


Problem 114
===========

A row measuring seven units in length has red blocks with a minimum length
of three units placed on it, such that any two red blocks (which are
allowed to be different lengths) are separated by at least one black
square. There are exactly seventeen ways of doing this.

                       +------+  +------+  +------+
                       +------+  +------+  +------+
                       +------+  +------+  +------+
                       +------+  +------+  +------+
                       +------+  +------+  +------+
                       +------+  +------+  +------+
                       +------+  +------+  +------+
                       +------+  +------+  +------+
                       +------+  +------+  +------+
                       +------+  +------+  +------+
                       +------+  +------+
                       +------+  +------+

How many ways can a row measuring fifty units in length be filled?

NOTE: Although the example above does not lend itself to the possibility,
in general it is permitted to mix block sizes. For example, on a row
measuring eight units in length you could use red (3), black (1), and red
(4).


Problem 115
===========

NOTE: This is a more difficult version of problem 114.

A row measuring n units in length has red blocks with a minimum length of
m units placed on it, such that any two red blocks (which are allowed to
be different lengths) are separated by at least one black square.

Let the fill-count function, F(m, n), represent the number of ways that a
row can be filled.

For example, F(3, 29) = 673135 and F(3, 30) = 1089155.

That is, for m = 3, it can be seen that n = 30 is the smallest value for
which the fill-count function first exceeds one million.

In the same way, for m = 10, it can be verified that F(10, 56) = 880711
and F(10, 57) = 1148904, so n = 57 is the least value for which the
fill-count function first exceeds one million.

For m = 50, find the least value of n for which the fill-count function
first exceeds one million.


Problem 116
===========

A row of five black square tiles is to have a number of its tiles replaced
with coloured oblong tiles chosen from red (length two), green (length
three), or blue (length four).

If red tiles are chosen there are exactly seven ways this can be done.

                      +----+  +----+  +----+  +----+
                      +----+  +----+  +----+  +----+

                      +----+  +----+  +----+
                      +----+  +----+  +----+

If green tiles are chosen there are three ways.

                        +----+  +----+  +----+
                        +----+  +----+  +----+

And if blue tiles are chosen there are two ways.

                              +----+  +----+
                              +----+  +----+

Assuming that colours cannot be mixed there are 7 + 3 + 2 = 12 ways of
replacing the black tiles in a row measuring five units in length.

How many different ways can the black tiles in a row measuring fifty units
in length be replaced if colours cannot be mixed and at least one coloured
tile must be used?

NOTE: This is related to problem 117.


Problem 117
===========

Using a combination of black square tiles and oblong tiles chosen from:
red tiles measuring two units, green tiles measuring three units, and blue
tiles measuring four units, it is possible to tile a row measuring five
units in length in exactly fifteen different ways.

                      +----+  +----+  +----+  +----+
                      +----+  +----+  +----+  +----+

                      +----+  +----+  +----+  +----+
                      +----+  +----+  +----+  +----+

                      +----+  +----+  +----+  +----+
                      +----+  +----+  +----+  +----+

                      +----+  +----+  +----+
                      +----+  +----+  +----+

How many ways can a row measuring fifty units in length be tiled?

NOTE: This is related to problem 116.


Problem 118
===========

Using all of the digits 1 through 9 and concatenating them freely to form
decimal integers, different sets can be formed. Interestingly with the set
{2,5,47,89,631}, all of the elements belonging to it are prime.

How many distinct sets containing each of the digits one through nine
exactly once contain only prime elements?


Problem 119
===========

The number 512 is interesting because it is equal to the sum of its digits
raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512. Another example of a
number with this property is 614656 = 28^4.

We shall define a[n] to be the nth term of this sequence and insist that a
number must contain at least two digits to have a sum.

You are given that a[2] = 512 and a[10] = 614656.

Find a[30].


Problem 120
===========

Let r be the remainder when (a1)^n + (a+1)^n is divided by a^2.

For example, if a = 7 and n = 3, then r = 42: 6^3 + 8^3 = 728 42 mod 49.
And as n varies, so too will r, but for a = 7 it turns out that r[max] =
42.

For 3 a 1000, find r[max].


Problem 121
===========

A bag contains one red disc and one blue disc. In a game of chance a
player takes a disc at random and its colour is noted. After each turn the
disc is returned to the bag, an extra red disc is added, and another disc
is taken at random.

The player pays -L-1 to play and wins if they have taken more blue discs
than red discs at the end of the game.

If the game is played for four turns, the probability of a player winning
is exactly 11/120, and so the maximum prize fund the banker should
allocate for winning in this game would be -L-10 before they would expect
to incur a loss. Note that any payout will be a whole number of pounds and
also includes the original -L-1 paid to play the game, so in the example
given the player actually wins -L-9.

Find the maximum prize fund that should be allocated to a single game in
which fifteen turns are played.


Problem 122
===========

The most naive way of computing n^15 requires fourteen multiplications:

n * n * ... * n = n^15

But using a "binary" method you can compute it in six multiplications:

n * n = n^2
n^2 * n^2 = n^4
n^4 * n^4 = n^8
n^8 * n^4 = n^12
n^12 * n^2 = n^14
n^14 * n = n^15

However it is yet possible to compute it in only five multiplications:

n * n = n^2
n^2 * n = n^3
n^3 * n^3 = n^6
n^6 * n^6 = n^12
n^12 * n^3 = n^15

We shall define m(k) to be the minimum number of multiplications to
compute n^k; for example m(15) = 5.

For 1 k 200, find m(k).


Problem 123
===========

Let p[n] be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder
when (p[n]1)^n + (p[n]+1)^n is divided by p[n]^2.

For example, when n = 3, p[3] = 5, and 4^3 + 6^3 = 280 5 mod 25.

The least value of n for which the remainder first exceeds 10^9 is 7037.

Find the least value of n for which the remainder first exceeds 10^10.


Problem 124
===========

The radical of n, rad(n), is the product of distinct prime factors of n.
For example, 504 = 2^3 * 3^2 * 7, so rad(504) = 2 * 3 * 7 = 42.

If we calculate rad(n) for 1 n 10, then sort them on rad(n), and sorting
on n if the radical values are equal, we get:

                         Unsorted       Sorted
                         n  rad(n)   n  rad(n) k
                         1    1      1    1    1
                         2    2      2    2    2
                         3    3      4    2    3
                         4    2      8    2    4
                         5    5      3    3    5
                         6    6      9    3    6
                         7    7      5    5    7
                         8    2      6    6    8
                         9    3      7    7    9
                         10   10     10   10   10

Let E(k) be the kth element in the sorted n column; for example, E(4) = 8
and E(6) = 9.

If rad(n) is sorted for 1 n 100000, find E(10000).


Problem 125
===========

The palindromic number 595 is interesting because it can be written as the
sum of consecutive squares: 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2.

There are exactly eleven palindromes below one-thousand that can be
written as consecutive square sums, and the sum of these palindromes is
4164. Note that 1 = 0^2 + 1^2 has not been included as this problem is
concerned with the squares of positive integers.

Find the sum of all the numbers less than 10^8 that are both palindromic
and can be written as the sum of consecutive squares.


Problem 126
===========

The minimum number of cubes to cover every visible face on a cuboid
measuring 3 x 2 x 1 is twenty-two.

If we then add a second layer to this solid it would require forty-six
cubes to cover every visible face, the third layer would require
seventy-eight cubes, and the fourth layer would require one-hundred and
eighteen cubes to cover every visible face.

However, the first layer on a cuboid measuring 5 x 1 x 1 also requires
twenty-two cubes; similarly the first layer on cuboids measuring
5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes.

We shall define C(n) to represent the number of solids that contain n
cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118)
= 8.

It turns out that 154 is the least value of n for which C(n) = 10.

Find the least value of n for which C(n) = 1000.


Problem 127
===========

The radical of n, rad(n), is the product of distinct prime factors of n.
For example, 504 = 2^3 * 3^2 * 7, so rad(504) = 2 * 3 * 7 = 42.

We shall define the triplet of positive integers (a, b, c) to be an
abc-hit if:

 1. GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
 2. a < b
 3. a + b = c
 4. rad(abc) < c

For example, (5, 27, 32) is an abc-hit, because:

 1. GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
 2. 5 < 27
 3. 5 + 27 = 32
 4. rad(4320) = 30 < 32

It turns out that abc-hits are quite rare and there are only thirty-one
abc-hits for c < 1000, with c = 12523.

Find c for c < 110000.


Problem 128
===========

A hexagonal tile with number 1 is surrounded by a ring of six hexagonal
tiles, starting at "12 o'clock" and numbering the tiles 2 to 7 in an
anti-clockwise direction.

New rings are added in the same fashion, with the next rings being
numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows
the first three rings.

By finding the difference between tile n and each its six neighbours we
shall define PD(n) to be the number of those differences which are prime.

For example, working clockwise around tile 8 the differences are 12, 29,
11, 6, 1, and 13. So PD(8) = 3.

In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and
10, hence PD(17) = 2.

It can be shown that the maximum value of PD(n) is 3.

If all of the tiles for which PD(n) = 3 are listed in ascending order to
form a sequence, the 10th tile would be 271.

Find the 2000th tile in this sequence.


Problem 129
===========

A number consisting entirely of ones is called a repunit. We shall define
R(k) to be a repunit of length k; for example, R(6) = 111111.

Given that n is a positive integer and GCD(n, 10) = 1, it can be shown
that there always exists a value, k, for which R(k) is divisible by n, and
let A(n) be the least such value of k; for example, A(7) = 6 and A(41) =
5.

The least value of n for which A(n) first exceeds ten is 17.

Find the least value of n for which A(n) first exceeds one-million.


Problem 130
===========

A number consisting entirely of ones is called a repunit. We shall define
R(k) to be a repunit of length k; for example, R(6) = 111111.

Given that n is a positive integer and GCD(n, 10) = 1, it can be shown
that there always exists a value, k, for which R(k) is divisible by n, and
let A(n) be the least such value of k; for example, A(7) = 6 and A(41) =
5.

You are given that for all primes, p > 5, that p 1 is divisible by A(p).
For example, when p = 41, A(41) = 5, and 40 is divisible by 5.

However, there are rare composite values for which this is also true; the
first five examples being 91, 259, 451, 481, and 703.

Find the sum of the first twenty-five composite values of n for which
GCD(n, 10) = 1 and n 1 is divisible by A(n).


Problem 131
===========

There are some prime values, p, for which there exists a positive integer,
n, such that the expression n^3 + n^2p is a perfect cube.

For example, when p = 19, 8^3 + 8^2 * 19 = 12^3.

What is perhaps most surprising is that for each prime with this property
the value of n is unique, and there are only four such primes below
one-hundred.

How many primes below one million have this remarkable property?


Problem 132
===========

A number consisting entirely of ones is called a repunit. We shall define
R(k) to be a repunit of length k.

For example, R(10) = 1111111111 = 11 * 41 * 271 * 9091, and the sum of
these prime factors is 9414.

Find the sum of the first forty prime factors of R(10^9).


Problem 133
===========

A number consisting entirely of ones is called a repunit. We shall define
R(k) to be a repunit of length k; for example, R(6) = 111111.

Let us consider repunits of the form R(10^n).

Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is
divisible by 17. Yet there is no value of n for which R(10^n) will divide
by 19. In fact, it is remarkable that 11, 17, 41, and 73 are only four
primes below one-hundred that can ever be a factor of R(10^n).

Find the sum of all the primes below one-hundred thousand that will never
be a factor of R(10^n).


Problem 134
===========

Consider the consecutive primes p[1] = 19 and p[2] = 23. It can be
verified that 1219 is the smallest number such that the last digits are
formed by p[1] whilst also being divisible by p[2].

In fact, with the exception of p[1] = 3 and p[2] = 5, for every pair of
consecutive primes, p[2] > p[1], there exist values of n for which the
last digits are formed by p[1] and n is divisible by p[2]. Let S be the
smallest of these values of n.

Find S for every pair of consecutive primes with 5 p[1] 1000000.


Problem 135
===========

Given the positive integers, x, y, and z, are consecutive terms of an
arithmetic progression, the least value of the positive integer, n, for
which the equation, x^2 y^2 z^2 = n, has exactly two solutions is n = 27:

                    34^2 27^2 20^2 = 12^2 9^2 6^2 = 27

It turns out that n = 1155 is the least value which has exactly ten
solutions.

How many values of n less than one million have exactly ten distinct
solutions?


Problem 136
===========

The positive integers, x, y, and z, are consecutive terms of an arithmetic
progression. Given that n is a positive integer, the equation, x^2 y^2 z^2
= n, has exactly one solution when n = 20:

                            13^2 10^2 7^2 = 20

In fact there are twenty-five values of n below one hundred for which the
equation has a unique solution.

How many values of n less than fifty million have exactly one solution?


Problem 137
===========

Consider the infinite polynomial series A[F](x) = xF[1] + x^2F[2] +
x^3F[3] + ..., where F[k] is the kth term in the Fibonacci sequence: 1, 1,
2, 3, 5, 8, ... ; that is, F[k] = F[k[1]] + F[k[2]], F[1] = 1 and F[2] =
1.

For this problem we shall be interested in values of x for which A[F](x)
is a positive integer.

Surprisingly A[F](1/2)  =  (1/2).1 + (1/2)^2.1 + (1/2)^3.2 + (1/2)^4.3 +
                           (1/2)^5.5 + ...
                        =  1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ...
                        =  2

The corresponding values of x for the first five natural numbers are shown
below.

                            +---------------+
                            |x      |A[F](x)|
                            |-------+-------|
                            |21     |1      |
                            |-------+-------|
                            |1/2    |2      |
                            |-------+-------|
                            |(132)/3|3      |
                            |-------+-------|
                            |(895)/8|4      |
                            |-------+-------|
                            |(343)/5|5      |
                            +---------------+

We shall call A[F](x) a golden nugget if x is rational, because they
become increasingly rarer; for example, the 10th golden nugget is
74049690.

Find the 15th golden nugget.


Problem 138
===========

Consider the isosceles triangle with base length, b = 16, and legs, L =
17.

By using the Pythagorean theorem it can be seen that the height of the
triangle, h = (17^2 8^2) = 15, which is one less than the base length.

With b = 272 and L = 305, we get h = 273, which is one more than the base
length, and this is the second smallest isosceles triangle with the
property that h = b 1.

Find L for the twelve smallest isosceles triangles for which h = b 1 and
b, L are positive integers.


Problem 139
===========

Let (a, b, c) represent the three sides of a right angle triangle with
integral length sides. It is possible to place four such triangles
together to form a square with length c.

For example, (3, 4, 5) triangles can be placed together to form a 5 by 5
square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5
square can be tiled with twenty-five 1 by 1 squares.

However, if (5, 12, 13) triangles were used then the hole would measure 7
by 7 and these could not be used to tile the 13 by 13 square.

Given that the perimeter of the right triangle is less than one-hundred
million, how many Pythagorean triangles would allow such a tiling to take
place?


Problem 140
===========

Consider the infinite polynomial series A[G](x) = xG[1] + x^2G[2] +
x^3G[3] + ..., where G[k] is the kth term of the second order recurrence
relation G[k] = G[k[1]] + G[k[2]], G[1] = 1 and G[2] = 4; that is, 1, 4,
5, 9, 14, 23, ... .

For this problem we shall be concerned with values of x for which A[G](x)
is a positive integer.

The corresponding values of x for the first five natural numbers are shown
below.

                           +-----------------+
                           |x        |A[G](x)|
                           |---------+-------|
                           |(51)/4   |1      |
                           |---------+-------|
                           |2/5      |2      |
                           |---------+-------|
                           |(222)/6  |3      |
                           |---------+-------|
                           |(1375)/14|4      |
                           |---------+-------|
                           |1/2      |5      |
                           +-----------------+

We shall call A[G](x) a golden nugget if x is rational, because they
become increasingly rarer; for example, the 20th golden nugget is
211345365.

Find the sum of the first thirty golden nuggets.


Problem 141
===========

A positive integer, n, is divided by d and the quotient and remainder are
q and r respectively. In addition d, q, and r are consecutive positive
integer terms in a geometric sequence, but not necessarily in that order.

For example, 58 divided by 6 has quotient 9 and remainder 4. It can also
be seen that 4, 6, 9 are consecutive terms in a geometric sequence (common
ratio 3/2).
We will call such numbers, n, progressive.

Some progressive numbers, such as 9 and 10404 = 102^2, happen to also be
perfect squares.
The sum of all progressive perfect squares below one hundred thousand is
124657.

Find the sum of all progressive perfect squares below one trillion
(10^12).


Problem 142
===========

Find the smallest x + y + z with integers x > y > z > 0 such that x + y, x
y, x + z, x z, y + z, y z are all perfect squares.


Problem 143
===========

Let ABC be a triangle with all interior angles being less than 120
degrees. Let X be any point inside the triangle and let XA = p, XB = q,
and XC = r.

Fermat challenged Torricelli to find the position of X such that p + q + r
was minimised.

Torricelli was able to prove that if equilateral triangles AOB, BNC and
AMC are constructed on each side of triangle ABC, the circumscribed
circles of AOB, BNC, and AMC will intersect at a single point, T, inside
the triangle. Moreover he proved that T, called the Torricelli/Fermat
point, minimises p + q + r. Even more remarkable, it can be shown that
when the sum is minimised, AN = BM = CO = p + q + r and that AN, BM and CO
also intersect at T.

If the sum is minimised and a, b, c, p, q and r are all positive integers
we shall call triangle ABC a Torricelli triangle. For example, a = 399, b
= 455, c = 511 is an example of a Torricelli triangle, with p + q + r =
784.

Find the sum of all distinct values of p + q + r 110000 for Torricelli
triangles.


Problem 144
===========

In laser physics, a "white cell" is a mirror system that acts as a delay
line for the laser beam. The beam enters the cell, bounces around on the
mirrors, and eventually works its way back out.

The specific white cell we will be considering is an ellipse with the
equation 4x^2 + y^2 = 100

The section corresponding to 0.01 x +0.01 at the top is missing, allowing
the light to enter and exit through the hole.

The light beam in this problem starts at the point (0.0,10.1) just outside
the white cell, and the beam first impacts the mirror at (1.4,-9.6).

Each time the laser beam hits the surface of the ellipse, it follows the
usual law of reflection "angle of incidence equals angle of reflection."
That is, both the incident and reflected beams make the same angle with
the normal line at the point of incidence.

In the figure on the left, the red line shows the first two points of
contact between the laser beam and the wall of the white cell; the blue
line shows the line tangent to the ellipse at the point of incidence of
the first bounce.

The slope m of the tangent line at any point (x,y) of the given ellipse
is: m = 4x/y

The normal line is perpendicular to this tangent line at the point of
incidence.

The animation on the right shows the first 10 reflections of the beam.

How many times does the beam hit the internal surface of the white cell
before exiting?


Problem 145
===========

Some positive integers n have the property that the sum [ n + reverse(n) ]
consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and
409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409,
and 904 are reversible. Leading zeroes are not allowed in either n or
reverse(n).

There are 120 reversible numbers below one-thousand.

How many reversible numbers are there below one-billion (10^9)?


Problem 146
===========

The smallest positive integer n for which the numbers n^2+1, n^2+3, n^2+7,
n^2+9, n^2+13, and n^2+27 are consecutive primes is 10. The sum of all
such integers n below one-million is 1242490.

What is the sum of all such integers n below 150 million?


Problem 147
===========

In a 3x2 cross-hatched grid, a total of 37 different rectangles could be
situated within that grid as indicated in the sketch.

There are 5 grids smaller than 3x2, vertical and horizontal dimensions
being important, i.e. 1x1, 2x1, 3x1, 1x2 and 2x2. If each of them is
cross-hatched, the following number of different rectangles could be
situated within those smaller grids:

1x1: 1
2x1: 4
3x1: 8
1x2: 4
2x2: 18

Adding those to the 37 of the 3x2 grid, a total of 72 different rectangles
could be situated within 3x2 and smaller grids.

How many different rectangles could be situated within 47x43 and smaller
grids?


Problem 148
===========

We can easily verify that none of the entries in the first seven rows of
Pascal's triangle are divisible by 7:

                                    1
                                 1     1
                              1     2     1
                           1     3     3     1
                        1     4     6     4     1
                     1     5    10    10     5     1
                  1     6    15    20    15     6     1

However, if we check the first one hundred rows, we will find that only
2361 of the 5050 entries are not divisible by 7.

Find the number of entries which are not divisible by 7 in the first one
billion (10^9) rows of Pascal's triangle.


Problem 149
===========

Looking at the table below, it is easy to verify that the maximum possible
sum of adjacent numbers in any direction (horizontal, vertical, diagonal
or anti-diagonal) is 16 (= 8 + 7 + 1).

                           +-----------------+
                           | 2 | 5 | 3 | 2   |
                           |---+---+---+-----|
                           | 9 | 6 | 5 | 1   |
                           |---+---+---+-----|
                           | 3 | 2 | 7 | 3   |
                           |---+---+---+-----|
                           | 1 | 8 | 4 |   8 |
                           +-----------------+

Now, let us repeat the search, but on a much larger scale:

First, generate four million pseudo-random numbers using a specific form
of what is known as a "Lagged Fibonacci Generator":

For 1 k 55, s[k] = [100003 200003k + 300007k^3] (modulo 1000000) 500000.
For 56 k 4000000, s[k] = [s[k24] + s[k55] + 1000000] (modulo 1000000)
500000.

Thus, s[10] = 393027 and s[100] = 86613.

The terms of s are then arranged in a 2000 * 2000 table, using the first
2000 numbers to fill the first row (sequentially), the next 2000 numbers
to fill the second row, and so on.

Finally, find the greatest sum of (any number of) adjacent entries in any
direction (horizontal, vertical, diagonal or anti-diagonal).


Problem 150
===========

In a triangular array of positive and negative integers, we wish to find a
sub-triangle such that the sum of the numbers it contains is the smallest
possible.

In the example below, it can be easily verified that the marked triangle
satisfies this condition having a sum of 42.

We wish to make such a triangular array with one thousand rows, so we
generate 500500 pseudo-random numbers s[k] in the range 2^19, using a type
of random number generator (known as a Linear Congruential Generator) as
follows:

t := 0
for k = 1 up to k = 500500:
    t := (615949*t + 797807) modulo 2^20
    s[k] := t2^19

Thus: s[1] = 273519, s[2] = 153582, s[3] = 450905 etc

Our triangular array is then formed using the pseudo-random numbers thus:

                                   s[1
                                ]s[2]  s[3
                            ]s[4]  s[5]  s[6]
                          s[7]  s[8]  s[9]  s[10
                                   ]...

Sub-triangles can start at any element of the array and extend down as far
as we like (taking-in the two elements directly below it from the next
row, the three elements directly below from the row after that, and so
on).
The "sum of a sub-triangle" is defined as the sum of all the elements it
contains.
Find the smallest possible sub-triangle sum.


Problem 151
===========

A printing shop runs 16 batches (jobs) every week and each batch requires
a sheet of special colour-proofing paper of size A5.

Every Monday morning, the foreman opens a new envelope, containing a large
sheet of the special paper with size A1.

He proceeds to cut it in half, thus getting two sheets of size A2. Then he
cuts one of them in half to get two sheets of size A3 and so on until he
obtains the A5-size sheet needed for the first batch of the week.

All the unused sheets are placed back in the envelope.

At the beginning of each subsequent batch, he takes from the envelope one
sheet of paper at random. If it is of size A5, he uses it. If it is
larger, he repeats the 'cut-in-half' procedure until he has what he needs
and any remaining sheets are always placed back in the envelope.

Excluding the first and last batch of the week, find the expected number
of times (during each week) that the foreman finds a single sheet of paper
in the envelope.

Give your answer rounded to six decimal places using the format x.xxxxxx .


Problem 152
===========

There are several ways to write the number 1/2 as a sum of inverse squares
using distinct integers.

For instance, the numbers {2,3,4,5,7,12,15,20,28,35} can be used:

In fact, only using integers between 2 and 45 inclusive, there are exactly
three ways to do it, the remaining two being:
{2,3,4,6,7,9,10,20,28,35,36,45} and {2,3,4,6,7,9,12,15,28,30,35,36,45}.

How many ways are there to write the number 1/2 as a sum of inverse
squares using distinct integers between 2 and 80 inclusive?


Problem 153
===========

As we all know the equation x^2=-1 has no solutions for real x.
If we however introduce the imaginary number i this equation has two
solutions: x=i and x=-i.
If we go a step further the equation (x-3)^2=-4 has two complex solutions:
x=3+2i and x=3-2i.
x=3+2i and x=3-2i are called each others' complex conjugate.
Numbers of the form a+bi are called complex numbers.
In general a+bi and abi are each other's complex conjugate.

A Gaussian Integer is a complex number a+bi such that both a and b are
integers.
The regular integers are also Gaussian integers (with b=0).
To distinguish them from Gaussian integers with b 0 we call such integers
"rational integers."
A Gaussian integer is called a divisor of a rational integer n if the
result is also a Gaussian integer.
If for example we divide 5 by 1+2i we can simplify in the following
manner:
Multiply numerator and denominator by the complex conjugate of 1+2i: 12i.
The result is .
So 1+2i is a divisor of 5.
Note that 1+i is not a divisor of 5 because .
Note also that if the Gaussian Integer (a+bi) is a divisor of a rational
integer n, then its complex conjugate (abi) is also a divisor of n.

In fact, 5 has six divisors such that the real part is positive: {1, 1 +
2i, 1 2i, 2 + i, 2 i, 5}.
The following is a table of all of the divisors for the first five
positive rational integers:

           +--------------------------------------------------+
           | n | Gaussian integer divisors    | Sum s(n) of   |
           |   | with positive real part      | thesedivisors |
           |---+------------------------------+---------------|
           | 1 | 1                            | 1             |
           |---+------------------------------+---------------|
           | 2 | 1, 1+i, 1-i, 2               | 5             |
           |---+------------------------------+---------------|
           | 3 | 1, 3                         | 4             |
           |---+------------------------------+---------------|
           | 4 | 1, 1+i, 1-i, 2, 2+2i, 2-2i,4 | 13            |
           |---+------------------------------+---------------|
           | 5 | 1, 1+2i, 1-2i, 2+i, 2-i, 5   | 12            |
           +--------------------------------------------------+

For divisors with positive real parts, then, we have: .

For 1 n 10^5, s(n)=17924657155.

What is s(n) for 1 n 10^8?


Problem 154
===========

A triangular pyramid is constructed using spherical balls so that each
ball rests on exactly three balls of the next lower level.

Then, we calculate the number of paths leading from the apex to each
position:

A path starts at the apex and progresses downwards to any of the three
spheres directly below the current position.

Consequently, the number of paths to reach a certain position is the sum
of the numbers immediately above it (depending on the position, there are
up to three numbers above it).

The result is Pascal's pyramid and the numbers at each level n are the
coefficients of the trinomial expansion (x + y + z)^n.

How many coefficients in the expansion of (x + y + z)^200000 are multiples
of 10^12?


Problem 155
===========

An electric circuit uses exclusively identical capacitors of the same
value C.
The capacitors can be connected in series or in parallel to form
sub-units, which can then be connected in series or in parallel with other
capacitors or other sub-units to form larger sub-units, and so on up to a
final circuit.

Using this simple procedure and up to n identical capacitors, we can make
circuits having a range of different total capacitances. For example,
using up to n=3 capacitors of 60 F each, we can obtain the following 7
distinct total capacitance values:

If we denote by D(n) the number of distinct total capacitance values we
can obtain when using up to n equal-valued capacitors and the simple
procedure described above, we have: D(1)=1, D(2)=3, D(3)=7 ...

Find D(18).

Reminder : When connecting capacitors C[1], C[2] etc in parallel, the
total capacitance is C[T] = C[1] + C[2] +...,
whereas when connecting them in series, the overall capacitance is given
by:


Problem 156
===========

Starting from zero the natural numbers are written down in base 10 like
this:
0 1 2 3 4 5 6 7 8 9 10 11 12....

Consider the digit d=1. After we write down each number n, we will update
the number of ones that have occurred and call this number f(n,1). The
first values for f(n,1), then, are as follows:

                                n  f(n,1)
                                0  0
                                1  1
                                2  1
                                3  1
                                4  1
                                5  1
                                6  1
                                7  1
                                8  1
                                9  1
                                10 2
                                11 4
                                12 5

Note that f(n,1) never equals 3.
So the first two solutions of the equation f(n,1)=n are n=0 and n=1. The
next solution is n=199981.

In the same manner the function f(n,d) gives the total number of digits d
that have been written down after the number n has been written.
In fact, for every digit d 0, 0 is the first solution of the equation
f(n,d)=n.

Let s(d) be the sum of all the solutions for which f(n,d)=n.
You are given that s(1)=22786974071.

Find s(d) for 1 d 9.

Note: if, for some n, f(n,d)=n for more than one value of d this value of
n is counted again for every value of d for which f(n,d)=n.


Problem 157
===========

Consider the diophantine equation ^1/[a]+^1/[b]= ^p/[10^n] with a, b, p, n
positive integers and a b.
For n=1 this equation has 20 solutions that are listed below:

/[1]+^1/[1]=^20/[10   ]^1/[1]+^1/[2]=^15/[10  ]^1/[1]+^1/[5]=^12/[10  ]^1/[1]+^1/[10]=^11/[10 ]^1/[2]+^1/[2]=^10/[10
/[2]+^1/[5]=^7/[10    ]^1/[2]+^1/[10]=^6/[10  ]^1/[3]+^1/[6]=^5/[10   ]^1/[3]+^1/[15]=^4/[10  ]^1/[4]+^1/[4]=^5/[10
/[4]+^1/[20]=^3/[10   ]^1/[5]+^1/[5]=^4/[10   ]^1/[5]+^1/[10]=^3/[10  ]^1/[6]+^1/[30]=^2/[10  ]^1/[10]+^1/[10]=^2/[10
/[11]+^1/[110]=^1/[10 ]^1/[12]+^1/[60]=^1/[10 ]^1/[14]+^1/[35]=^1/[10 ]^1/[15]+^1/[30]=^1/[10 ]^1/[20]+^1/[20]=^1/[10

]How many solutions has this equation for 1 n 9?


Problem 158
===========

Taking three different letters from the 26 letters of the alphabet,
character strings of length three can be formed.
Examples are 'abc', 'hat' and 'zyx'.
When we study these three examples we see that for 'abc' two characters
come lexicographically after its neighbour to the left.
For 'hat' there is exactly one character that comes lexicographically
after its neighbour to the left. For 'zyx' there are zero characters that
come lexicographically after its neighbour to the left.
In all there are 10400 strings of length 3 for which exactly one character
comes lexicographically after its neighbour to the left.

We now consider strings of n 26 different characters from the alphabet.
For every n, p(n) is the number of strings of length n for which exactly
one character comes lexicographically after its neighbour to the left.

What is the maximum value of p(n)?


Problem 159
===========

A composite number can be factored many different ways. For instance, not
including multiplication by one, 24 can be factored in 7 distinct ways:

24 = 2x2x2x3
24 = 2x3x4
24 = 2x2x6
24 = 4x6
24 = 3x8
24 = 2x12
24 = 24

Recall that the digital root of a number, in base 10, is found by adding
together the digits of that number, and repeating that process until a
number is arrived at that is less than 10. Thus the digital root of 467 is
8.

We shall call a Digital Root Sum (DRS) the sum of the digital roots of the
individual factors of our number.
The chart below demonstrates all of the DRS values for 24.

                     +------------------------------+
                     |Factorisation|Digital Root Sum|
                     |-------------+----------------|
                     |2x2x2x3      |       9        |
                     |-------------+----------------|
                     |2x3x4        |       9        |
                     |-------------+----------------|
                     |2x2x6        |       10       |
                     |-------------+----------------|
                     |4x6          |       10       |
                     |-------------+----------------|
                     |3x8          |       11       |
                     |-------------+----------------|
                     |2x12         |       5        |
                     |-------------+----------------|
                     |24           |       6        |
                     +------------------------------+

The maximum Digital Root Sum of 24 is 11.
The function mdrs(n) gives the maximum Digital Root Sum of n. So
mdrs(24)=11.
Find mdrs(n) for 1 < n < 1,000,000.


Problem 160
===========

For any N, let f(N) be the last five digits before the trailing zeroes in
N!.
For example,

9! = 362880 so f(9)=36288
10! = 3628800 so f(10)=36288
20! = 2432902008176640000 so f(20)=17664

Find f(1,000,000,000,000)


Problem 161
===========

A triomino is a shape consisting of three squares joined via the
edges.There are two basic forms:

If all possible orientations are taken into account there are six:

Any n by m grid for which nxm is divisible by 3 can be tiled with
triominoes.
If we consider tilings that can be obtained by reflection or rotation from
another tiling as different there are 41 ways a 2 by 9 grid can be tiled
with triominoes:

In how many ways can a 9 by 12 grid be tiled in this way by triominoes?


Problem 162
===========

In the hexadecimal number system numbers are represented using 16
different digits:

                     0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

The hexadecimal number AF when written in the decimal number system equals
10x16+15=175.

In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1
and A are all present.
Like numbers written in base ten we write hexadecimal numbers without
leading zeroes.

How many hexadecimal numbers containing at most sixteen hexadecimal digits
exist with all of the digits 0,1, and A present at least once?
Give your answer as a hexadecimal number.

(A,B,C,D,E and F in upper case, without any leading or trailing code that
marks the number as hexadecimal and without leading zeroes , e.g. 1A3F and
not: 1a3f and not 0x1a3f and not $1A3F and not #1A3F and not 0000001A3F)


Problem 163
===========

Consider an equilateral triangle in which straight lines are drawn from
each vertex to the middle of the opposite side, such as in the size 1
triangle in the sketch below.

Sixteen triangles of either different shape or size or orientation or
location can now be observed in that triangle. Using size 1 triangles as
building blocks, larger triangles can be formed, such as the size 2
triangle in the above sketch. One-hundred and four triangles of either
different shape or size or orientation or location can now be observed in
that size 2 triangle.

It can be observed that the size 2 triangle contains 4 size 1 triangle
building blocks. A size 3 triangle would contain 9 size 1 triangle
building blocks and a size n triangle would thus contain n^2 size 1
triangle building blocks.

If we denote T(n) as the number of triangles present in a triangle of size
n, then

T(1) = 16
T(2) = 104

Find T(36).


Problem 164
===========

How many 20 digit numbers n (without any leading zero) exist such that no
three consecutive digits of n have a sum greater than 9?


Problem 165
===========

A segment is uniquely defined by its two endpoints.
By considering two line segments in plane geometry there are three
possibilities:
the segments have zero points, one point, or infinitely many points in
common.

Moreover when two segments have exactly one point in common it might be
the case that that common point is an endpoint of either one of the
segments or of both. If a common point of two segments is not an endpoint
of either of the segments it is an interior point of both segments.
We will call a common point T of two segments L[1] and L[2] a true
intersection point of L[1] and L[2] if T is the only common point of L[1]
and L[2] and T is an interior point of both segments.

Consider the three segments L[1], L[2], and L[3]:

L[1]: (27, 44) to (12, 32)
L[2]: (46, 53) to (17, 62)
L[3]: (46, 70) to (22, 40)

It can be verified that line segments L[2] and L[3] have a true
intersection point. We note that as the one of the end points of L[3]:
(22,40) lies on L[1] this is not considered to be a true point of
intersection. L[1] and L[2] have no common point. So among the three line
segments, we find one true intersection point.

Now let us do the same for 5000 line segments. To this end, we generate
20000 numbers using the so-called "Blum Blum Shub" pseudo-random number
generator.

s[0] = 290797

s[n+1] = s[n] * s[n] (modulo 50515093)

t[n] = s[n] (modulo 500)

To create each line segment, we use four consecutive numbers t[n]. That
is, the first line segment is given by:

(t[1], t[2]) to (t[3], t[4])

The first four numbers computed according to the above generator should
be: 27, 144, 12 and 232. The first segment would thus be (27,144) to
(12,232).

How many distinct true intersection points are found among the 5000 line
segments?


Problem 166
===========

A 4x4 grid is filled with digits d, 0 d 9.

It can be seen that in the grid

                                 6 3 3 0
                                 5 0 4 3
                                 0 7 1 4
                                 1 2 4 5

the sum of each row and each column has the value 12. Moreover the sum of
each diagonal is also 12.

In how many ways can you fill a 4x4 grid with the digits d, 0 d 9 so that
each row, each column, and both diagonals have the same sum?


Problem 167
===========

For two positive integers a and b, the Ulam sequence U(a,b) is defined by
U(a,b)[1] = a, U(a,b)[2] = b and for k > 2,U(a,b)[k] is the smallest
integer greater than U(a,b)[(k-1)] which can be written in exactly one way
as the sum of two distinct previous members of U(a,b).

For example, the sequence U(1,2) begins with
1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8;
5 does not belong to it because 5 = 1 + 4 = 2 + 3 has two representations
as the sum of two previous members, likewise 7 = 1 + 6 = 3 + 4.

Find U(2,2n+1)[k] for 2 n 10, where k = 10^11.


Problem 168
===========

Consider the number 142857. We can right-rotate this number by moving the
last digit (7) to the front of it, giving us 714285.
It can be verified that 714285=5 * 142857.
This demonstrates an unusual property of 142857: it is a divisor of its
right-rotation.

Find the last 5 digits of the sum of all integers n, 10 < n < 10^100, that
have this property.


Problem 169
===========

Define f(0)=1 and f(n) to be the number of different ways n can be
expressed as a sum of integer powers of 2 using each power no more than
twice.

For example, f(10)=5 since there are five different ways to express 10:

1 + 1 + 8
1 + 1 + 4 + 4
1 + 1 + 2 + 2 + 4
2 + 4 + 4
2 + 8

What is f(10^25)?


Problem 170
===========

Take the number 6 and multiply it by each of 1273 and 9854:

6 * 1273 = 7638
6 * 9854 = 59124

By concatenating these products we get the 1 to 9 pandigital 763859124. We
will call 763859124 the "concatenated product of 6 and (1273,9854)".
Notice too, that the concatenation of the input numbers, 612739854, is
also 1 to 9 pandigital.

The same can be done for 0 to 9 pandigital numbers.

What is the largest 0 to 9 pandigital 10-digit concatenated product of an
integer with two or more other integers, such that the concatenation of
the input numbers is also a 0 to 9 pandigital 10-digit number?


Problem 171
===========

For a positive integer n, let f(n) be the sum of the squares of the digits
(in base 10) of n, e.g.

f(3) = 3^2 = 9,
f(25) = 2^2 + 5^2 = 4 + 25 = 29,
f(442) = 4^2 + 4^2 + 2^2 = 16 + 16 + 4 = 36

Find the last nine digits of the sum of all n, 0 < n < 10^20, such that
f(n) is a perfect square.


Problem 172
===========

How many 18-digit numbers n (without leading zeros) are there such that no
digit occurs more than three times in n?


Problem 173
===========

We shall define a square lamina to be a square outline with a square
"hole" so that the shape possesses vertical and horizontal symmetry. For
example, using exactly thirty-two square tiles we can form two different
square laminae:

With one-hundred tiles, and not necessarily using all of the tiles at one
time, it is possible to form forty-one different square laminae.

Using up to one million tiles how many different square laminae can be
formed?


Problem 174
===========

We shall define a square lamina to be a square outline with a square
"hole" so that the shape possesses vertical and horizontal symmetry.

Given eight tiles it is possible to form a lamina in only one way: 3x3
square with a 1x1 hole in the middle. However, using thirty-two tiles it
is possible to form two distinct laminae.

If t represents the number of tiles used, we shall say that t = 8 is type
L(1) and t = 32 is type L(2).

Let N(n) be the number of t 1000000 such that t is type L(n); for example,
N(15) = 832.

What is N(n) for 1 n 10?


Problem 175
===========

Define f(0)=1 and f(n) to be the number of ways to write n as a sum of
powers of 2 where no power occurs more than twice.

For example, f(10)=5 since there are five different ways to express 10:
10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1

It can be shown that for every fraction p/q (p > 0, q > 0) there exists at
least one integer n such that
f(n)/f(n-1)=p/q.

For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241.
The binary expansion of 241 is 11110001.
Reading this binary number from the most significant bit to the least
significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the
string 4,3,1 the Shortened Binary Expansion of 241.

Find the Shortened Binary Expansion of the smallest n for which
f(n)/f(n-1)=123456789/987654321.

Give your answer as comma separated integers, without any whitespaces.



Problem 176
===========

The four rectangular triangles with sides (9,12,15), (12,16,20), (5,12,13)
and (12,35,37) all have one of the shorter sides (catheti) equal to 12. It
can be shown that no other integer sided rectangular triangle exists with
one of the catheti equal to 12.

Find the smallest integer that can be the length of a cathetus of exactly
47547 different integer sided rectangular triangles.


Problem 177
===========

Let ABCD be a convex quadrilateral, with diagonals AC and BD. At each
vertex the diagonal makes an angle with each of the two sides, creating
eight corner angles.

For example, at vertex A, the two angles are CAD, CAB.

We call such a quadrilateral for which all eight corner angles have
integer values when measured in degrees an "integer angled quadrilateral".
An example of an integer angled quadrilateral is a square, where all eight
corner angles are 45DEG. Another example is given by DAC = 20DEG, BAC =
60DEG, ABD = 50DEG, CBD = 30DEG, BCA = 40DEG, DCA = 30DEG, CDB = 80DEG,
ADB = 50DEG.

What is the total number of non-similar integer angled quadrilaterals?

Note: In your calculations you may assume that a calculated angle is
integral if it is within a tolerance of 10^-9 of an integer value.


Problem 178
===========

Consider the number 45656.
It can be seen that each pair of consecutive digits of 45656 has a
difference of one.
A number for which every pair of consecutive digits has a difference of
one is called a step number.
A pandigital number contains every decimal digit from 0 to 9 at least
once.
How many pandigital step numbers less than 10^40 are there?



Problem 179
===========

Find the number of integers 1 < n < 10^7, for which n and n + 1 have the
same number of positive divisors. For example, 14 has the positive
divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15.


Problem 180
===========

For any integer n, consider the three functions

f[1,n](x,y,z) = x^n+1 + y^n+1 z^n+1
f[2,n](x,y,z) = (xy + yz + zx)*(x^n-1 + y^n-1 z^n-1)
f[3,n](x,y,z) = xyz*(x^n-2 + y^n-2 z^n-2)

and their combination

f[n](x,y,z) = f[1,n](x,y,z) + f[2,n](x,y,z) f[3,n](x,y,z)

We call (x,y,z) a golden triple of order k if x, y, and z are all rational
numbers of the form a / b with
0 < a < b k and there is (at least) one integer n, so that f[n](x,y,z) =
0.

Let s(x,y,z) = x + y + z.
Let t = u / v be the sum of all distinct s(x,y,z) for all golden triples
(x,y,z) of order 35.
All the s(x,y,z) and t must be in reduced form.

Find u + v.


Problem 181
===========

Having three black objects B and one white object W they can be grouped in
7 ways like this:

     (BBBW)  (B,BBW)  (B,B,BW)  (B,B,B,W)  (B,BB,W)  (BBB,W)  (BB,BW)

In how many ways can sixty black objects B and forty white objects W be
thus grouped?


Problem 182
===========

The RSA encryption is based on the following procedure:

Generate two distinct primes p and q.
Compute n=pq and f=(p-1)(q-1).
Find an integer e, 1 < e < f, such that gcd(e,f)=1.

A message in this system is a number in the interval [0,n-1].
A text to be encrypted is then somehow converted to messages (numbers in
the interval [0,n-1]).
To encrypt the text, for each message, m, c=m^e mod n is calculated.

To decrypt the text, the following procedure is needed: calculate d such
that ed=1 mod f, then for each encrypted message, c, calculate m=c^d mod
n.

There exist values of e and m such that m^e mod n=m.
We call messages m for which m^e mod n=m unconcealed messages.

An issue when choosing e is that there should not be too many unconcealed
messages.
For instance, let p=19 and q=37.
Then n=19*37=703 and f=18*36=648.
If we choose e=181, then, although gcd(181,648)=1 it turns out that all
possible messages
m (0mn-1) are unconcealed when calculating m^e mod n.
For any valid choice of e there exist some unconcealed messages.
It's important that the number of unconcealed messages is at a minimum.

Choose p=1009 and q=3643.
Find the sum of all values of e, 1 < e < f(1009,3643) and gcd(e,f)=1, so
that the number of unconcealed messages for this value of e is at a
minimum.


Problem 183
===========

Let N be a positive integer and let N be split into k equal parts, r =
N/k, so that N = r + r + ... + r.
Let P be the product of these parts, P = r * r * ... * r = r^k.

For example, if 11 is split into five equal parts, 11 = 2.2 + 2.2 + 2.2 +
2.2 + 2.2, then P = 2.2^5 = 51.53632.

Let M(N) = P[max] for a given value of N.

It turns out that the maximum for N = 11 is found by splitting eleven into
four equal parts which leads to P[max] = (11/4)^4; that is, M(11) =
14641/256 = 57.19140625, which is a terminating decimal.

However, for N = 8 the maximum is achieved by splitting it into three
equal parts, so M(8) = 512/27, which is a non-terminating decimal.

Let D(N) = N if M(N) is a non-terminating decimal and D(N) = -N if M(N) is
a terminating decimal.

For example, SD(N) for 5 N 100 is 2438.

Find SD(N) for 5 N 10000.


Problem 184
===========

Consider the set I[r] of points (x,y) with integer co-ordinates in the
interior of the circle with radius r, centered at the origin, i.e. x^2 +
y^2 < r^2.

For a radius of 2, I[2] contains the nine points (0,0), (1,0), (1,1),
(0,1), (-1,1), (-1,0), (-1,-1), (0,-1) and (1,-1). There are eight
triangles having all three vertices in I[2] which contain the origin in
the interior. Two of them are shown below, the others are obtained from
these by rotation.

For a radius of 3, there are 360 triangles containing the origin in the
interior and having all vertices in I[3] and for I[5] the number is 10600.

How many triangles are there containing the origin in the interior and
having all three vertices in I[105]?


Problem 185
===========

The game Number Mind is a variant of the well known game Master Mind.

Instead of coloured pegs, you have to guess a secret sequence of digits.
After each guess you're only told in how many places you've guessed the
correct digit. So, if the sequence was 1234 and you guessed 2036, you'd be
told that you have one correct digit; however, you would NOT be told that
you also have another digit in the wrong place.

For instance, given the following guesses for a 5-digit secret sequence,

90342 ;2 correct
70794 ;0 correct
39458 ;2 correct
34109 ;1 correct
51545 ;2 correct
12531 ;1 correct

The correct sequence 39542 is unique.

Based on the following guesses,

5616185650518293 ;2 correct
3847439647293047 ;1 correct
5855462940810587 ;3 correct
9742855507068353 ;3 correct
4296849643607543 ;3 correct
3174248439465858 ;1 correct
4513559094146117 ;2 correct
7890971548908067 ;3 correct
8157356344118483 ;1 correct
2615250744386899 ;2 correct
8690095851526254 ;3 correct
6375711915077050 ;1 correct
6913859173121360 ;1 correct
6442889055042768 ;2 correct
2321386104303845 ;0 correct
2326509471271448 ;2 correct
5251583379644322 ;2 correct
1748270476758276 ;3 correct
4895722652190306 ;1 correct
3041631117224635 ;3 correct
1841236454324589 ;3 correct
2659862637316867 ;2 correct

Find the unique 16-digit secret sequence.


Problem 186
===========

Here are the records from a busy telephone system with one million users:

                 +-------------------------------------+
                 |RecNr            | Caller  | Called  |
                 |-----------------+---------+---------|
                 |        1        | 200007  | 100053  |
                 |-----------------+---------+---------|
                 |        2        | 600183  | 500439  |
                 |-----------------+---------+---------|
                 |        3        | 600863  | 701497  |
                 |-----------------+---------+---------|
                 |       ...       |   ...   |   ...   |
                 +-------------------------------------+

The telephone number of the caller and the called number in record n are
Caller(n) = S[2n-1] and Called(n) = S[2n] where S[1,2,3,...] come from the
"Lagged Fibonacci Generator":

For 1 k 55, S[k] = [100003 - 200003k + 300007k^3] (modulo 1000000)
For 56 k, S[k] = [S[k-24] + S[k-55]] (modulo 1000000)

If Caller(n) = Called(n) then the user is assumed to have misdialled and
the call fails; otherwise the call is successful.

From the start of the records, we say that any pair of users X and Y are
friends if X calls Y or vice-versa. Similarly, X is a friend of a friend
of Z if X is a friend of Y and Y is a friend of Z; and so on for longer
chains.

The Prime Minister's phone number is 524287. After how many successful
calls, not counting misdials, will 99% of the users (including the PM) be
a friend, or a friend of a friend etc., of the Prime Minister?


Problem 187
===========

A composite is a number containing at least two prime factors. For
example, 15 = 3 * 5; 9 = 3 * 3; 12 = 2 * 2 * 3.

There are ten composites below thirty containing precisely two, not
necessarily distinct, prime factors:4, 6, 9, 10, 14, 15, 21, 22, 25, 26.

How many composite integers, n < 10^8, have precisely two, not necessarily
distinct, prime factors?


Problem 188
===========

The hyperexponentiation or tetration of a number a by a positive integer
b, denoted by a-^-^b or ^ba, is recursively defined by:

a-^-^1 = a,
a-^-^(k+1) = a^(a-^-^k).

Thus we have e.g. 3-^-^2 = 3^3 = 27, hence 3-^-^3 = 3^27 = 7625597484987
and 3-^-^4 is roughly 10^3.6383346400240996*10^12.

Find the last 8 digits of 1777-^-^1855.


Problem 189
===========

Consider the following configuration of 64 triangles:

We wish to colour the interior of each triangle with one of three colours:
red, green or blue, so that no two neighbouring triangles have the same
colour. Such a colouring shall be called valid. Here, two triangles are
said to be neighbouring if they share an edge.
Note: if they only share a vertex, then they are not neighbours.

For example, here is a valid colouring of the above grid:

A colouring C' which is obtained from a colouring C by rotation or
reflection is considered distinct from C unless the two are identical.

How many distinct valid colourings are there for the above configuration?


Problem 190
===========

Let S[m] = (x[1], x[2], ... , x[m]) be the m-tuple of positive real
numbers with x[1] + x[2] + ... + x[m] = m for which P[m] = x[1] * x[2]^2 *
... * x[m]^m is maximised.

For example, it can be verified that [P[10]] = 4112 ([ ] is the integer
part function).

Find S[P[m]] for 2 m 15.


Problem 191
===========

A particular school offers cash rewards to children with good attendance
and punctuality. If they are absent for three consecutive days or late on
more than one occasion then they forfeit their prize.

During an n-day period a trinary string is formed for each child
consisting of L's (late), O's (on time), and A's (absent).

Although there are eighty-one trinary strings for a 4-day period that can
be formed, exactly forty-three strings would lead to a prize:

OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA
OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO
AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL
AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA
LAOO LAOA LAAO

How many "prize" strings exist over a 30-day period?


Problem 192
===========

Let x be a real number.
A best approximation to x for the denominator bound d is a rational number
r/s in reduced form, with s d, such that any rational number which is
closer to x than r/s has a denominator larger than d:

                         |p/q-x| < |r/s-x| q > d

For example, the best approximation to 13 for the denominator bound 20 is
18/5 and the best approximation to 13 for the denominator bound 30 is
101/28.

Find the sum of all denominators of the best approximations to n for the
denominator bound 10^12, where n is not a perfect square and 1 < n 100000.


Problem 193
===========

A positive integer n is called squarefree, if no square of a prime divides
n, thus 1, 2, 3, 5, 6, 7, 10, 11 are squarefree, but not 4, 8, 9, 12.

How many squarefree numbers are there below 2^50?


Problem 194
===========

Consider graphs built with the units A: and B: , where the units are glued
alongthe vertical edges as in the graph .

A configuration of type (a,b,c) is a graph thus built of a units A and b
units B, where the graph's vertices are coloured using up to c colours, so
that no two adjacent vertices have the same colour.
The compound graph above is an example of a configuration of type (2,2,6),
in fact of type (2,2,c) for all c 4.

Let N(a,b,c) be the number of configurations of type (a,b,c).
For example, N(1,0,3) = 24, N(0,2,4) = 92928 and N(2,2,3) = 20736.

Find the last 8 digits of N(25,75,1984).


Problem 195
===========

Let's call an integer sided triangle with exactly one angle of 60 degrees
a 60-degree triangle.
Let r be the radius of the inscribed circle of such a 60-degree triangle.

There are 1234 60-degree triangles for which r 100.
Let T(n) be the number of 60-degree triangles for which r n, so
T(100) = 1234,  T(1000) = 22767, and  T(10000) = 359912.

Find T(1053779).


Problem 196
===========

Build a triangle from all positive integers in the following way:

 1
 2  3
 4  5  6
 7  8  9 10
11 12 13 14 15
16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63 64 65 66
. . .

Each positive integer has up to eight neighbours in the triangle.

A set of three primes is called a prime triplet if one of the three primes
has the other two as neighbours in the triangle.

For example, in the second row, the prime numbers 2 and 3 are elements of
some prime triplet.

If row 8 is considered, it contains two primes which are elements of some
prime triplet, i.e. 29 and 31.
If row 9 is considered, it contains only one prime which is an element of
some prime triplet: 37.

Define S(n) as the sum of the primes in row n which are elements of any
prime triplet.
Then S(8)=60 and S(9)=37.

You are given that S(10000)=950007619.

Find  S(5678027) + S(7208785).


Problem 197
===========

Given is the function f(x) = 2^30.403243784-x^2 * 10^-9 ( is the
floor-function),
the sequence u[n] is defined by u[0] = -1 and u[n+1] = f(u[n]).

Find u[n] + u[n+1] for n = 10^12.
Give your answer with 9 digits after the decimal point.


Problem 198
===========

A best approximation to a real number x for the denominator bound d is a
rational number r/s (in reduced form) with s d, so that any rational
number p/q which is closer to x than r/s has q > d.

Usually the best approximation to a real number is uniquely determined for
all denominator bounds. However, there are some exceptions, e.g. 9/40 has
the two best approximations 1/4 and 1/5 for the denominator bound 6.We
shall call a real number x ambiguous, if there is at least one denominator
bound for which x possesses two best approximations. Clearly, an ambiguous
number is necessarily rational.

How many ambiguous numbers x = p/q,0 < x < 1/100, are there whose
denominator q does not exceed 10^8?


Problem 199
===========

Three circles of equal radius are placed inside a larger circle such that
each pair of circles is tangent to one another and the inner circles do
not overlap. There are four uncovered "gaps" which are to be filled
iteratively with more tangent circles.

At each iteration, a maximally sized circle is placed in each gap, which
creates more gaps for the next iteration. After 3 iterations (pictured),
there are 108 gaps and the fraction of the area which is not covered by
circles is 0.06790342, rounded to eight decimal places.

What fraction of the area is not covered by circles after 10 iterations?
Give your answer rounded to eight decimal places using the format
x.xxxxxxxx .


Problem 200
===========

We shall define a sqube to be a number of the form, p^2q^3, where p and q
are distinct primes.
For example, 200 = 5^22^3 or 120072949 = 23^261^3.

The first five squbes are 72, 108, 200, 392, and 500.

Interestingly, 200 is also the first number for which you cannot change
any single digit to make a prime; we shall call such numbers, prime-proof.
The next prime-proof sqube which contains the contiguous sub-string "200"
is 1992008.

Find the 200th prime-proof sqube containing the contiguous sub-string
"200".


Problem 201
===========

For any set A of numbers, let sum(A) be the sum of the elements of A.
Consider the set B = {1,3,6,8,10,11}.
There are 20 subsets of B containing three elements, and their sums are:

sum({1,3,6}) = 10,
sum({1,3,8}) = 12,
sum({1,3,10}) = 14,
sum({1,3,11}) = 15,
sum({1,6,8}) = 15,
sum({1,6,10}) = 17,
sum({1,6,11}) = 18,
sum({1,8,10}) = 19,
sum({1,8,11}) = 20,
sum({1,10,11}) = 22,
sum({3,6,8}) = 17,
sum({3,6,10}) = 19,
sum({3,6,11}) = 20,
sum({3,8,10}) = 21,
sum({3,8,11}) = 22,
sum({3,10,11}) = 24,
sum({6,8,10}) = 24,
sum({6,8,11}) = 25,
sum({6,10,11}) = 27,
sum({8,10,11}) = 29.

Some of these sums occur more than once, others are unique.
For a set A, let U(A,k) be the set of unique sums of k-element subsets of
A, in our example we find U(B,3) = {10,12,14,18,21,25,27,29} and
sum(U(B,3)) = 156.

Now consider the 100-element set S = {1^2, 2^2, ... , 100^2}.
S has 100891344545564193334812497256 50-element subsets.

Determine the sum of all integers which are the sum of exactly one of the
50-element subsets of S, i.e. find sum(U(S,50)).


Problem 202
===========

Three mirrors are arranged in the shape of an equilateral triangle, with
their reflective surfaces pointing inwards. There is an infinitesimal gap
at each vertex of the triangle through which a laser beam may pass.

Label the vertices A, B and C. There are 2 ways in which a laser beam may
enter vertex C, bounce off 11 surfaces, then exit through the same vertex:
one way is shown below; the other is the reverse of that.

[Image: 202_laserbeam.gif]

There are 80840 ways in which a laser beam may enter vertex C, bounce off
1000001 surfaces, then exit through the same vertex.

In how many ways can a laser beam enter at vertex C, bounce off
12017639147 surfaces, then exit through the same vertex?


Problem 203
===========

The binomial coefficients ^nC[k] can be arranged in triangular form,
Pascal's triangle, like this:

                                   1
                                1     1
                             1     2     1
                          1     3     3     1
                        1    4     6     4     1
                      1   5     10    10    5    1
                    1   6    15    20    15    6   1
                  1   7   21    35    35    21   7   1

                               .........

It can be seen that the first eight rows of Pascal's triangle contain
twelve distinct numbers: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 and 35.

A positive integer n is called squarefree if no square of a prime divides
n.Of the twelve distinct numbers in the first eight rows of Pascal's
triangle, all except 4 and 20 are squarefree.The sum of the distinct
squarefree numbers in the first eight rows is 105.

Find the sum of the distinct squarefree numbers in the first 51 rows of
Pascal's triangle.


Problem 204
===========

A Hamming number is a positive number which has no prime factor larger
than 5.
So the first few Hamming numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15.
There are 1105 Hamming numbers not exceeding 10^8.

We will call a positive number a generalised Hamming number of type n, if
it has no prime factor larger than n.
Hence the Hamming numbers are the generalised Hamming numbers of type 5.

How many generalised Hamming numbers of type 100 are there which don't
exceed 10^9?


Problem 205
===========

Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2,
3, 4.
Colin has six six-sided (cubic) dice, each with faces numbered 1, 2, 3, 4,
5, 6.

Peter and Colin roll their dice and compare totals: the highest total
wins. The result is a draw if the totals are equal.

What is the probability that Pyramidal Pete beats Cubic Colin? Give your
answer rounded to seven decimal places in the form 0.abcdefg


Problem 206
===========

Find the unique positive integer whose square has the form
1_2_3_4_5_6_7_8_9_0, where each “_” is a single digit.


Problem 207
===========

For some positive integers k, there exists an integer partition of the
form 4^t = 2^t + k, where 4^t, 2^t, and k are all positive integers and
t is a real number.

The first two such partitions are 4^1 = 2^1 + 2 and 4^1.5849625... =
2^1.5849625... + 6.

Partitions where t is also an integer are called perfect.
For any m ≥ 1 let P(m) be the proportion of such partitions that are
perfect with k ≤ m.
Thus P(6) = 1/2.

In the following table are listed some values of P(m)

   P(5) = 1/1
   P(10) = 1/2
   P(15) = 2/3
   P(20) = 1/2
   P(25) = 1/2
   P(30) = 2/5
   ...
   P(180) = 1/4
   P(185) = 3/13

Find the smallest m for which P(m) < 1/12345


Problem 208
===========

A robot moves in a series of one-fifth circular arcs (72°), with a free
choice of a clockwise or an anticlockwise arc for each step, but no
turning on the spot.

One of 70932 possible closed paths of 25 arcs starting northward is:

[Image: 208_robotwalk.gif]

Given that the robot starts facing North, how many journeys of 70 arcs in
length can it take that return it, after the final arc, to its starting
position?
(Any arc may be traversed multiple times.)


Problem 209
===========

A k-input binary truth table is a map from k input bits(binary digits, 0
[false] or 1 [true]) to 1 output bit. For example, the 2-input binary
truth tables for the logical AND and XOR functions are:

            ┌────┬────┬─────────┐        ┌────┬────┬─────────┐
            │ x  │ y  │ x AND y │        │ x  │ y  │ x XOR y │
            ├────┼────┼─────────┤        ├────┼────┼─────────┤
            │ 0  │ 0  │    0    │        │ 0  │ 0  │    0    │
            ├────┼────┼─────────┤        ├────┼────┼─────────┤
            │ 0  │ 1  │    0    │        │ 0  │ 1  │    1    │
            ├────┼────┼─────────┤        ├────┼────┼─────────┤
            │ 1  │ 0  │    0    │        │ 1  │ 0  │    1    │
            ├────┼────┼─────────┤        ├────┼────┼─────────┤
            │ 1  │ 1  │    1    │        │ 1  │ 1  │    0    │
            └────┴────┴─────────┘        └────┴────┴─────────┘

How many 6-input binary truth tables, τ, satisfy the formula

     τ(a, b, c, d, e, f) AND τ(b, c, d, e, f, a XOR (b AND c)) = 0


Problem 210
===========

   Consider the set S(r) of points (x,y) with integer coordinates satisfying
   |x| + |y| ≤ r.
   Let O be the point (0,0) and C the point (r/4,r/4).
   Let N(r) be the number of points B in S(r), so that the triangle OBC has
   an obtuse angle, i.e. the largest angle α satisfies 90°<α<180°.
   So, for example, N(4)=24 and N(8)=100.

   What is N(1,000,000,000)?


Problem 211
===========

For a positive integer n, let σ[2](n) be the sum of the squares of its
divisors. For example,

                   σ[2](10) = 1 + 4 + 25 + 100 = 130.

Find the sum of all n, 0 < n < 64,000,000 such that σ[2](n) is a perfect
square.


Problem 212
===========

An axis-aligned cuboid, specified by parameters { (x[0],y[0],z[0]),
(dx,dy,dz) }, consists of all points (X,Y,Z) such that x[0] ≤ X ≤ x[0]+dx,
y[0] ≤ Y ≤ y[0]+dy and z[0] ≤ Z ≤ z[0]+dz. The volume of the cuboid is the
product, dx × dy × dz. The combined volume of a collection of cuboids is
the volume of their union and will be less than the sum of the individual
volumes if any cuboids overlap.

Let C[1],...,C[50000] be a collection of 50000 axis-aligned cuboids such
that C[n] has parameters

x[0] = S[6n-5] modulo 10000
y[0] = S[6n-4] modulo 10000
z[0] = S[6n-3] modulo 10000
dx = 1 + (S[6n-2] modulo 399)
dy = 1 + (S[6n-1] modulo 399)
dz = 1 + (S[6n] modulo 399)

where S[1],...,S[300000] come from the "Lagged Fibonacci Generator":

For 1 ≤ k ≤ 55, S[k] = [100003 - 200003k + 300007k^3]   (modulo 1000000)
For 56 ≤ k, S[k] = [S[k-24] + S[k-55]]   (modulo 1000000)

Thus, C[1] has parameters {(7,53,183),(94,369,56)}, C[2] has parameters
{(2383,3563,5079),(42,212,344)}, and so on.

The combined volume of the first 100 cuboids, C[1],...,C[100], is
723581599.

What is the combined volume of all 50000 cuboids, C[1],...,C[50000]?


Problem 213
===========

A 30×30 grid of squares contains 900 fleas, initially one flea per square.
When a bell is rung, each flea jumps to an adjacent square at random
(usually 4 possibilities, except for fleas on the edge of the grid or at
the corners).

What is the expected number of unoccupied squares after 50 rings of the
bell? Give your answer rounded to six decimal places.


Problem 214
===========

Let φ be Euler's totient function, i.e. for a natural number n,φ(n) is the
number of k, 1 ≤ k ≤ n, for which gcd(k,n) = 1.

By iterating φ, each positive integer generates a decreasing chain of
numbers ending in 1.
E.g. if we start with 5 the sequence 5,4,2,1 is generated.
Here is a listing of all chains with length 4:

                                                                  5,4,2,1
                                                                  7,6,2,1
                                                                  8,4,2,1
                                                                  9,6,2,1
                                                                 10,4,2,1
                                                                 12,4,2,1
                                                                 14,6,2,1
                                                                 18,6,2,1

Only two of these chains start with a prime, their sum is 12.

What is the sum of all primes less than 40000000 which generate a chain of
length 25?


Problem 215
===========

Consider the problem of building a wall out of 2×1 and 3×1 bricks
(horizontal×vertical dimensions) such that, for extra strength, the gaps
between horizontally-adjacent bricks never line up in consecutive layers,
i.e. never form a "running crack".

For example, the following 9×3 wall is not acceptable due to the running
crack shown in red:

[Image: 215_crackfree.gif]

There are eight ways of forming a crack-free 9×3 wall, written W(9,3) = 8.

Calculate W(32,10).


Problem 216
===========

 Consider numbers t(n) of the form t(n) = 2n^2-1 with n > 1.
 The first such numbers are 7, 17, 31, 49, 71, 97, 127 and 161.
 It turns out that only 49 = 7*7 and 161 = 7*23 are not prime.
 For n ≤ 10000 there are 2202 numbers t(n) that are prime.

 How many numbers t(n) are prime for n ≤ 50,000,000?


Problem 217
===========

A positive integer with k (decimal) digits is called balanced if its first
⌈^k/[2]⌉ digits sum to the same value as its last ⌈^k/[2]⌉ digits, where
⌈x⌉, pronounced ceiling of x, is the smallest integer ≥ x, thus ⌈π⌉ = 4
and ⌈5⌉ = 5.

So, for example, all palindromes are balanced, as is 13722.

Let T(n) be the sum of all balanced numbers less than 10^n.
Thus: T(1) = 45, T(2) = 540 and T(5) = 334795890.

Find T(47) mod 3^15


Problem 218
===========

Consider the right angled triangle with sides a=7, b=24 and c=25.The area
of this triangle is 84, which is divisible by the perfect numbers 6 and
28.
Moreover it is a primitive right angled triangle as gcd(a,b)=1 and
gcd(b,c)=1.
Also c is a perfect square.

We will call a right angled triangle perfect if
-it is a primitive right angled triangle
-its hypotenuse is a perfect square

We will call a right angled triangle super-perfect if
-it is a perfect right angled triangle and
-its area is a multiple of the perfect numbers 6 and 28.

How many perfect right-angled triangles with c≤10^16 exist that are not
super-perfect?


Problem 219
===========

Let A and B be bit strings (sequences of 0's and 1's).
If A is equal to the leftmost length(A) bits of B, then A is said to be a
prefix of B.
For example, 00110 is a prefix of 001101001, but not of 00111 or 100110.

A prefix-free code of size n is a collection of n distinct bit strings
such that no string is a prefix of any other. For example, this is a
prefix-free code of size 6:

                      0000, 0001, 001, 01, 10, 11

Now suppose that it costs one penny to transmit a '0' bit, but four pence
to transmit a '1'.
Then the total cost of the prefix-free code shown above is 35 pence, which
happens to be the cheapest possible for the skewed pricing scheme in
question.
In short, we write Cost(6) = 35.

What is Cost(10^9) ?


Problem 220
===========

Let D[0] be the two-letter string "Fa". For n≥1, derive D[n] from D[n-1]
by the string-rewriting rules:

"a" → "aRbFR"
"b" → "LFaLb"

Thus, D[0] = "Fa", D[1] = "FaRbFR", D[2] = "FaRbFRRLFaLbFR", and so on.

These strings can be interpreted as instructions to a computer graphics
program, with "F" meaning "draw forward one unit", "L" meaning "turn left
90 degrees", "R" meaning "turn right 90 degrees", and "a" and "b" being
ignored. The initial position of the computer cursor is (0,0), pointing up
towards (0,1).

Then D[n] is an exotic drawing known as the Heighway Dragon of order n.
For example, D[10] is shown below; counting each "F" as one step, the
highlighted spot at (18,16) is the position reached after 500 steps.

[Image: 220.gif]

What is the position of the cursor after 10^12 steps in D[50]?
Give your answer in the form x,y with no spaces.


Problem 221
===========

We shall call a positive integer A an "Alexandrian integer", if there
exist integers p, q, r such that:

A = p · q · r    and   1 = 1 + 1 + 1
                      A   p   q   r

For example, 630 is an Alexandrian integer (p = 5, q = −7, r = −18).In
fact, 630 is the 6th Alexandrian integer, the first 6 Alexandrian
integers being: 6, 42, 120, 156, 420 and 630.

Find the 150000th Alexandrian integer.


Problem 222
===========

What is the length of the shortest pipe, of internal radius 50mm, that can
fully contain 21 balls of radii 30mm, 31mm, ..., 50mm?

Give your answer in micrometres (10^-6 m) rounded to the nearest integer.


Problem 223
===========

Let us call an integer sided triangle with sides a ≤ b ≤ c barely acute if
the sides satisfy a^2 + b^2 = c^2 + 1.

How many barely acute triangles are there with perimeter ≤ 25,000,000?


Problem 224
===========

Let us call an integer sided triangle with sides a ≤ b ≤ c barely obtuse
if the sides satisfy a^2 + b^2 = c^2 - 1.

How many barely obtuse triangles are there with perimeter ≤ 75,000,000?


Problem 225
===========

The sequence 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201 ...
is defined by T[1] = T[2] = T[3] = 1 and T[n] = T[n-1] + T[n-2] + T[n-3].

It can be shown that 27 does not divide any terms of this sequence.
In fact, 27 is the first odd number with this property.

Find the 124th odd number that does not divide any terms of the above
sequence.


Problem 226
===========

The blancmange curve is the set of points (x,y) such that 0 ≤ x ≤ 1 and

[Image: 226_formula.gif]

where s(x) = the distance from x to the nearest integer.

The area under the blancmange curve is equal to ½, shown in pink in the
diagram below.

[Image: 226_scoop2.gif]

Let C be the circle with centre (¼,½) and radius ¼, shown in black in the
diagram.

What area under the blancmange curve is enclosed by C?
Give your answer rounded to eight decimal places in the form 0.abcdefgh


Problem 227
===========

"The Chase" is a game played with two dice and an even number of players.

The players sit around a table; the game begins with two opposite players
having one die each. On each turn, the two players with a die roll it.
If a player rolls a 1, he passes the die to his neighbour on the left; if
he rolls a 6, he passes the die to his neighbour on the right; otherwise,
he keeps the die for the next turn.
The game ends when one player has both dice after they have been rolled
and passed; that player has then lost.

In a game with 100 players, what is the expected number of turns the game
lasts?

Give your answer rounded to ten significant digits.


Problem 228
===========

Let S[n] be the regular n-sided polygon – or shape – whose vertices v[k]
(k = 1,2,…,n) have coordinates:

      x[k]   =   cos( ^2k-1/[n] × 180° )
      y[k]   =   sin( ^2k-1/[n] × 180° )

Each S[n] is to be interpreted as a filled shape consisting of all points
on the perimeter and in the interior.

The Minkowski sum, S+T, of two shapes S and T is the result of adding
every point in S to every point in T, where point addition is performed
coordinate-wise: (u, v) + (x, y) = (u+x, v+y).

For example, the sum of S[3] and S[4] is the six-sided shape shown in pink
below:

[Image: 228.png]

How many sides does S[1864] + S[1865] + … + S[1909] have?


Problem 229
===========

Consider the number 3600. It is very special, because

3600 = 48^2 +   36^2

3600 = 20^2 + 2×40^2

3600 = 30^2 + 3×30^2

3600 = 45^2 + 7×15^2

Similarly, we find that 88201 = 99^2 + 280^2 = 287^2 + 2×54^2 = 283^2 +
3×52^2 = 197^2 + 7×84^2.

In 1747, Euler proved which numbers are representable as a sum of two
squares.We are interested in the numbers n which admit representations of
all of the following four types:

n = a[1]^2 +   b[1]^2

n = a[2]^2 + 2×b[2]^2

n = a[3]^2 + 3×b[3]^2

n = a[7]^2 + 7×b[7]^2,

where the a[k] and b[k] are positive integers.

There are 75373 such numbers that do not exceed 10^7.
How many such numbers are there that do not exceed 2×10^9?


Problem 230
===========

For any two strings of digits, A and B, we define F[A,B] to be the
sequence (A,B,AB,BAB,ABBAB,...) in which each term is the concatenation of
the previous two.

Further, we define D[A,B](n) to be the n-th digit in the first term of
F[A,B] that contains at least n digits.

Example:

Let A=1415926535, B=8979323846. We wish to find D[A,B](35), say.

The first few terms of F[A,B] are:
1415926535
8979323846
14159265358979323846
897932384614159265358979323846
14159265358979323846897932384614159265358979323846

Then D[A,B](35) is the 35th digit in the fifth term, which is 9.

Now we use for A the first 100 digits of π behind the decimal point:

14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679

and for B the next hundred digits:

82148086513282306647093844609550582231725359408128
48111745028410270193852110555964462294895493038196.

Find ∑[n = 0,1,...,17]   10^n× D[A,B]((127+19n)×7^n).


Problem 231
===========

The binomial coefficient ^10C[3] = 120.
120 = 2^3 × 3 × 5 = 2 × 2 × 2 × 3 × 5, and 2 + 2 + 2 + 3 + 5 = 14.
So the sum of the terms in the prime factorisation of ^10C[3] is 14.

Find the sum of the terms in the prime factorisation of
^20000000C[15000000].


Problem 232
===========

Two players share an unbiased coin and take it in turns to play "The
Race". On Player 1's turn, he tosses the coin once: if it comes up Heads,
he scores one point; if it comes up Tails, he scores nothing. On Player
2's turn, she chooses a positive integer T and tosses the coin T times: if
it comes up all Heads, she scores 2^T-1 points; otherwise, she scores
nothing. Player 1 goes first. The winner is the first to 100 or more
points.

On each turn Player 2 selects the number, T, of coin tosses that maximises
the probability of her winning.

What is the probability that Player 2 wins?

Give your answer rounded to eight decimal places in the form 0.abcdefgh.


Problem 233
===========

Let f(N) be the number of points with integer coordinates that are on a
circle passing through (0,0), (N,0),(0,N), and (N,N).

It can be shown that f(10000) = 36.

What is the sum of all positive integers N ≤ 10^11 such that f(N) = 420?


Problem 234
===========

For an integer n ≥ 4, we define the lower prime square root of n, denoted
by lps(n), as the largest prime ≤ √n and the upper prime square root of n,
ups(n), as the smallest prime ≥ √n.

So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37.
Let us call an integer n ≥ 4 semidivisible, if one of lps(n) and ups(n)
divides n, but not both.

The sum of the semidivisible numbers not exceeding 15 is 30, the numbers
are 8, 10 and 12.
15 is not semidivisible because it is a multiple of both lps(15) = 3 and
ups(15) = 5.
As a further example, the sum of the 92 semidivisible numbers up to 1000
is 34825.

What is the sum of all semidivisible numbers not exceeding 999966663333?


Problem 235
===========

Given is the arithmetic-geometric sequence u(k) = (900-3k)r^k-1.
Let s(n) = Σ[k=1...n]u(k).

Find the value of r for which s(5000) = -600,000,000,000.

Give your answer rounded to 12 places behind the decimal point.


Problem 236
===========

Suppliers 'A' and 'B' provided the following numbers of products for the
luxury hamper market:

                         Product       'A'  'B'
                    Beluga Caviar      5248 640
                    Christmas Cake     1312 1888
                    Gammon Joint       2624 3776
                    Vintage Port       5760 3776
                    Champagne Truffles 3936 5664

Although the suppliers try very hard to ship their goods in perfect
condition, there is inevitably some spoilage - i.e. products gone bad.

The suppliers compare their performance using two types of statistic:

 • The five per-product spoilage rates for each supplier are equal to the
   number of products gone bad divided by the number of products
   supplied, for each of the five products in turn.
 • The overall spoilage rate for each supplier is equal to the total
   number of products gone bad divided by the total number of products
   provided by that supplier.

To their surprise, the suppliers found that each of the five per-product
spoilage rates was worse (higher) for 'B' than for 'A' by the same factor
(ratio of spoilage rates), m>1; and yet, paradoxically, the overall
spoilage rate was worse for 'A' than for 'B', also by a factor of m.

There are thirty-five m>1 for which this surprising result could have
occurred, the smallest of which is 1476/1475.

What's the largest possible value of m?
Give your answer as a fraction reduced to its lowest terms, in the form
u/v.


Problem 237
===========

Let T(n) be the number of tours over a 4 × n playing board such that:

 • The tour starts in the top left corner.
 • The tour consists of moves that are up, down, left, or right one
   square.
 • The tour visits each square exactly once.
 • The tour ends in the bottom left corner.

The diagram shows one tour over a 4 × 10 board:

[Image: 237.gif]

T(10) is 2329. What is T(10^12) modulo 10^8?


Problem 238
===========

Create a sequence of numbers using the "Blum Blum Shub" pseudo-random
number generator:

                      s[0]   = 14025256
                      s[n+1] = s[n]^2 mod 20300713

Concatenate these numbers  s[0]s[1]s[2]… to create a string w of infinite
length.
Then, w = 14025256741014958470038053646…

For a positive integer k, if no substring of w exists with a sum of digits
equal to k, p(k) is defined to be zero. If at least one substring of w
exists with a sum of digits equal to k, we define p(k) = z, where z is the
starting position of the earliest such substring.

For instance:

The substrings 1, 14, 1402, …
with respective sums of digits equal to 1, 5, 7, …
start at position 1, hence p(1) = p(5) = p(7) = … = 1.

The substrings 4, 402, 4025, …
with respective sums of digits equal to 4, 6, 11, …
start at position 2, hence p(4) = p(6) = p(11) = … = 2.

The substrings 02, 0252, …
with respective sums of digits equal to 2, 9, …
start at position 3, hence p(2) = p(9) = … = 3.

Note that substring 025 starting at position 3, has a sum of digits equal
to 7, but there was an earlier substring (starting at position 1) with a
sum of digits equal to 7, so p(7) = 1, not 3.

We can verify that, for 0 < k ≤ 10^3, ∑ p(k) = 4742.

Find ∑ p(k), for 0 < k ≤ 2·10^15.


Problem 239
===========

A set of disks numbered 1 through 100 are placed in a line in random
order.

What is the probability that we have a partial derangement such that
exactly 22 prime number discs are found away from their natural positions?
(Any number of non-prime disks may also be found in or out of their
natural positions.)

Give your answer rounded to 12 places behind the decimal point in the form
0.abcdefghijkl.


Problem 240
===========

There are 1111 ways in which five 6-sided dice (sides numbered 1 to 6) can
be rolled so that the top three sum to 15. Some examples are:

D[1],D[2],D[3],D[4],D[5] = 4,3,6,3,5
D[1],D[2],D[3],D[4],D[5] = 4,3,3,5,6
D[1],D[2],D[3],D[4],D[5] = 3,3,3,6,6
D[1],D[2],D[3],D[4],D[5] = 6,6,3,3,3

In how many ways can twenty 12-sided dice (sides numbered 1 to 12) be
rolled so that the top ten sum to 70?


Problem 241
===========

For a positive integer n, let σ(n) be the sum of all divisors of n, so
e.g. σ(6) = 1 + 2 + 3 + 6 = 12.

A perfect number, as you probably know, is a number with σ(n) = 2n.

Let us define the perfection quotient of a positive integer as
p(n) = σ(n) / n.

Find the sum of all positive integers n ≤ 10^18 for which p(n) has the
form k + ^1⁄[2], where k is an integer.


Problem 242
===========

Given the set {1,2,...,n}, we define f(n,k) as the number of its k-element
subsets with an odd sum of elements. For example, f(5,3) = 4, since the
set {1,2,3,4,5} has four 3-element subsets having an odd sum of elements,
i.e.: {1,2,4}, {1,3,5}, {2,3,4} and {2,4,5}.

When all three values n, k and f(n,k) are odd, we say that they make
an odd-triplet [n,k,f(n,k)].

There are exactly five odd-triplets with n ≤ 10, namely:
[1,1,f(1,1) = 1], [5,1,f(5,1) = 3], [5,5,f(5,5) = 1], [9,1,f(9,1) = 5] and
[9,9,f(9,9) = 1].

How many odd-triplets are there with n ≤ 10^12?


Problem 243
===========

A positive fraction whose numerator is less than its denominator is called
a proper fraction.
For any denominator, d, there will be d−1 proper fractions; for example,
with d = 12:
1/12 , 2/12 , 3/12 , 4/12 , 5/12 , 6/12 , 7/12 ,
8/12 , 9/12 , 10/12 , 11/12 .

We shall call a fraction that cannot be cancelled down a resilient
fraction.
Furthermore we shall define the resilience of a denominator, R(d), to be
the ratio of its proper fractions that are resilient; for example, R(12) =
4/11.
In fact, d = 12 is the smallest denominator having a resilience R(d) <
4/10.

Find the smallest denominator d, having a resilience R(d) < 15499/94744.


Problem 244
===========

You probably know the game Fifteen Puzzle. Here, instead of numbered
tiles, we have seven red tiles and eight blue tiles.

A move is denoted by the uppercase initial of the direction (Left, Right,
Up, Down) in which the tile is slid, e.g. starting from configuration (S),
by the sequence LULUR we reach the configuration (E):

(S): [Image: 244_start.gif]
(E): [Image: 244_example.gif]

For each path, its checksum is calculated by (pseudocode):
checksum = 0
checksum = (checksum × 243 + m[1]) mod 100 000 007
checksum = (checksum × 243 + m[2]) mod 100 000 007
   …
checksum = (checksum × 243 + m[n]) mod 100 000 007
where m[k] is the ASCII value of the k-th letter in the move sequence and
                  the ASCII values for the moves are:

                             ┌──────┬─────┐
                             │L     │76   │
                             ├──────┼─────┤
                             │R     │82   │
                             ├──────┼─────┤
                             │U     │85   │
                             ├──────┼─────┤
                             │D     │68   │
                             └──────┴─────┘

For the sequence LULUR given above, the checksum would be 19761398.

Now, starting from configuration (S), find all shortest ways to reach
configuration (T).

(S): [Image: 244_start.gif]
(T): [Image: 244_target.gif]

What is the sum of all checksums for the paths having the minimal length?


Problem 245
===========

We shall call a fraction that cannot be cancelled down a resilient
fraction. Furthermore we shall define the resilience of a denominator,
R(d), to be the ratio of its proper fractions that are resilient; for
example, R(12) = 4⁄/11.

The resilience of a number d > 1 is then φ(d) / (d - 1), where φ is Euler's
totient function.

We further define the coresilience of a number n > 1 as
C(n) = (n - φ(n)) / (n-1).

The coresilience of a prime p is C(p) = 1 / (p - 1).

Find the sum of all composite integers 1 < n ≤ 2×10^11, for which C(n) is
a unit fraction.


Problem 246
===========

A definition for an ellipse is:
Given a circle c with centre M and radius r and a point G such that
d(G,M)<r, the locus of the points that are equidistant from c and G form
an ellipse.

The construction of the points of the ellipse is shown below.

[Image: 246_anim.gif]

Given are the points M(-2000,1500) and G(8000,1500).
Given is also the circle c with centre M and radius 15000.
The locus of the points that are equidistant from G and c form an ellipse
e.
From a point P outside e the two tangents t[1] and t[2] to the ellipse are
drawn.
Let the points where t[1] and t[2] touch the ellipse be R and S.

[Image: 246_ellipse.gif]

For how many lattice points P is angle RPS greater than 45 degrees?


Problem 247
===========

Consider the region constrained by 1 ≤ x and 0 ≤ y ≤ 1/x.

Let S[1] be the largest square that can fit under the curve.
Let S[2] be the largest square that fits in the remaining area, and so on.
Let the index of S[n] be the pair (left, below) indicating the number of
squares to the left of S[n] and the number of squares below S[n].

[Image: 247_hypersquares.gif]

The diagram shows some such squares labelled by number.
S[2] has one square to its left and none below, so the index of S[2] is
(1,0).
It can be seen that the index of S[32] is (1,1) as is the index of S[50].
50 is the largest n for which the index of S[n] is (1,1).

What is the largest n for which the index of S[n] is (3,3)?


Problem 248
===========

The first number n for which φ(n)=13! is 6227180929.

Find the 150,000th such number.


Problem 249
===========

Let S = {2, 3, 5, ..., 4999} be the set of prime numbers less than 5000.

Find the number of subsets of S, the sum of whose elements is a prime
number.
Enter the rightmost 16 digits as your answer.


Problem 250
===========

Find the number of non-empty subsets of {1^1, 2^2, 3^3, ...,
250250^250250}, the sum of whose elements is divisible by 250. Enter the
rightmost 16 digits as your answer.


Problem 251
===========

A triplet of positive integers (a,b,c) is called a Cardano Triplet if it
satisfies the condition:

[Image: 251_cardano.gif]

For example, (2,1,5) is a Cardano Triplet.

There exist 149 Cardano Triplets for which a+b+c ≤ 1000.

Find how many Cardano Triplets exist such that a+b+c ≤ 110,000,000.


Problem 252
===========

Given a set of points on a plane, we define a convex hole to be a convex
polygon having as vertices any of the given points and not containing any
of the given points in its interior (in addition to the vertices, other
given points may lie on the perimeter of the polygon).

As an example, the image below shows a set of twenty points and a few such
convex holes. The convex hole shown as a red heptagon has an area equal to
1049694.5 square units, which is the highest possible area for a convex
hole on the given set of points.

[Image: 252_convexhole.gif]

For our example, we used the first 20 points (T[2k−1], T[2k]), for
k = 1,2,…,20, produced with the pseudo-random number generator:

                 S[0]   =    290797
                 S[n+1] =    S[n]^2 mod 50515093
                 T[n]   =    ( S[n] mod 2000 ) − 1000

i.e. (527, 144), (−488, 732), (−454, −947), …

What is the maximum area for a convex hole on the set containing the first
500 points in the pseudo-random sequence?
Specify your answer including one digit after the decimal point.


Problem 253
===========

A small child has a “number caterpillar” consisting of forty jigsaw
pieces, each with one number on it, which, when connected together in a
line, reveal the numbers 1 to 40 in order.

Every night, the child's father has to pick up the pieces of the
caterpillar that have been scattered across the play room. He picks up the
pieces at random and places them in the correct order.
As the caterpillar is built up in this way, it forms distinct segments
that gradually merge together.
The number of segments starts at zero (no pieces placed), generally
increases up to about eleven or twelve, then tends to drop again before
finishing at a single segment (all pieces placed).

For example:

                     ┌────────────┬───────────────┐
                     │Piece Placed│Segments So Far│
                     ├────────────┼───────────────┤
                     │     12     │       1       │
                     ├────────────┼───────────────┤
                     │     4      │       2       │
                     ├────────────┼───────────────┤
                     │     29     │       3       │
                     ├────────────┼───────────────┤
                     │     6      │       4       │
                     ├────────────┼───────────────┤
                     │     34     │       5       │
                     ├────────────┼───────────────┤
                     │     5      │       4       │
                     ├────────────┼───────────────┤
                     │     35     │       4       │
                     ├────────────┼───────────────┤
                     │     …      │       …       │
                     └────────────┴───────────────┘

Let M be the maximum number of segments encountered during a random
tidy-up of the caterpillar.
For a caterpillar of ten pieces, the number of possibilities for each M is

                        ┌────────┬─────────────┐
                        │   M    │Possibilities│
                        ├────────┼─────────────┤
                        │   1    │    512      │
                        ├────────┼─────────────┤
                        │   2    │ 250912      │
                        ├────────┼─────────────┤
                        │   3    │1815264      │
                        ├────────┼─────────────┤
                        │   4    │1418112      │
                        ├────────┼─────────────┤
                        │   5    │ 144000      │
                        └────────┴─────────────┘

so the most likely value of M is 3 and the average value is
^385643⁄[113400] = 3.400732, rounded to six decimal places.

The most likely value of M for a forty-piece caterpillar is 11; but what
is the average value of M?

Give your answer rounded to six decimal places.


Problem 254
===========

Define f(n) as the sum of the factorials of the digits of n. For example,
f(342) = 3! + 4! + 2! = 32.

Define sf(n) as the sum of the digits of f(n). So sf(342) = 3 + 2 = 5.

Define g(i) to be the smallest positive integer n such that sf(n) = i.
Though sf(342) is 5, sf(25) is also 5, and it can be verified that g(5) is
25.

Define sg(i) as the sum of the digits of g(i). So sg(5) = 2 + 5 = 7.

Further, it can be verified that g(20) is 267 and ∑ sg(i) for 1 ≤ i ≤ 20
is 156.

What is ∑ sg(i) for 1 ≤ i ≤ 150?


Problem 255
===========

We define the rounded-square-root of a positive integer n as the square
root of n rounded to the nearest integer.

The following procedure (essentially Heron's method adapted to integer
arithmetic) finds the rounded-square-root of n:

Let d be the number of digits of the number n.
If d is odd, set x[0] = 2×10^((d-1)⁄/2).
If d is even, set x[0] = 7×10^((d-2)⁄/2).
Repeat:

[Image: 255_Heron.gif]

until x[k+1] = x[k].

As an example, let us find the rounded-square-root of n = 4321.
n has 4 digits, so x[0] = 7×10^(4-2)⁄2 = 70.

[Image: 255_Example.gif]

Since x[2] = x[1], we stop here.
So, after just two iterations, we have found that the rounded-square-root
of 4321 is 66 (the actual square root is 65.7343137…).

The number of iterations required when using this method is surprisingly
low.
For example, we can find the rounded-square-root of a 5-digit integer
(10,000 ≤ n ≤ 99,999) with an average of 3.2102888889 iterations (the
average value was rounded to 10 decimal places).

Using the procedure described above, what is the average number of
iterations required to find the rounded-square-root of a 14-digit number
(10^13 ≤ n < 10^14)?
Give your answer rounded to 10 decimal places.

Note: The symbols ⌊x⌋ and ⌈x⌉ represent the floor function and ceiling
function respectively.


Problem 256
===========

Tatami are rectangular mats, used to completely cover the floor of a room,
without overlap.

Assuming that the only type of available tatami has dimensions 1×2, there
are obviously some limitations for the shape and size of the rooms that
can be covered.

For this problem, we consider only rectangular rooms with integer
dimensions a, b and even size s = a·b.
We use the term 'size' to denote the floor surface area of the room, and —
without loss of generality — we add the condition a ≤ b.

There is one rule to follow when laying out tatami: there must be no
points where corners of four different mats meet.
For example, consider the two arrangements below for a 4×4 room:

[Image: 256_tatami3.gif]

The arrangement on the left is acceptable, whereas the one on the right is
not: a red "X" in the middle, marks the point where four tatami meet.

Because of this rule, certain even-sized rooms cannot be covered with
tatami: we call them tatami-free rooms.
Further, we define T(s) as the number of tatami-free rooms of size s.

The smallest tatami-free room has size s = 70 and dimensions 7×10.
All the other rooms of size s = 70 can be covered with tatami; they are:
1×70, 2×35 and 5×14.
Hence, T(70) = 1.

Similarly, we can verify that T(1320) = 5 because there are exactly 5
tatami-free rooms of size s = 1320:
20×66, 22×60, 24×55, 30×44 and 33×40.
In fact, s = 1320 is the smallest room-size s for which T(s) = 5.

Find the smallest room-size s for which T(s) = 200.


Problem 257
===========

Given is an integer sided triangle ABC with sides a ≤ b ≤ c. (AB = c, BC =
a and AC = b).
The angular bisectors of the triangle intersect the sides at points E, F
and G (see picture below).

[Image: 257_bisector.gif]

The segments EF, EG and FG partition the triangle ABC into four smaller
triangles: AEG, BFE, CGF and EFG.
It can be proven that for each of these four triangles the ratio
area(ABC)/area(subtriangle) is rational.
However, there exist triangles for which some or all of these ratios are
integral.

How many triangles ABC with perimeter ≤ 100,000,000 exist so that the ratio
area(ABC)/area(AEG) is integral?


Problem 258
===========

A sequence is defined as:

 • g[k] = 1, for 0 ≤ k ≤ 1999
 • g[k] = g[k-2000] + g[k-1999], for k ≥ 2000.

Find g[k] mod 20092010 for k = 10^18.


Problem 259
===========

A positive integer will be called reachable if it can result from an
arithmetic expression obeying the following rules:

 • Uses the digits 1 through 9, in that order and exactly once each.
 • Any successive digits can be concatenated (for example, using the
   digits 2, 3 and 4 we obtain the number 234).
 • Only the four usual binary arithmetic operations (addition,
   subtraction, multiplication and division) are allowed.
 • Each operation can be used any number of times, or not at all.
 • Unary minus is not allowed.
 • Any number of (possibly nested) parentheses may be used to define the
   order of operations.

For example, 42 is reachable, since (1/23) * ((4*5)-6) * (78-9) = 42.

What is the sum of all positive reachable integers?


Problem 260
===========

A game is played with three piles of stones and two players.
At her turn, a player removes one or more stones from the piles. However,
if she takes stones from more than one pile, she must remove the same
number of stones from each of the selected piles.

In other words, the player chooses some N>0 and removes:

 • N stones from any single pile; or
 • N stones from each of any two piles (2N total); or
 • N stones from each of the three piles (3N total).

The player taking the last stone(s) wins the game.

A winning configuration is one where the first player can force a win.
For example, (0,0,13), (0,11,11) and (5,5,5) are winning configurations
because the first player can immediately remove all stones.

A losing configuration is one where the second player can force a win, no
matter what the first player does.
For example, (0,1,2) and (1,3,3) are losing configurations: any legal move
leaves a winning configuration for the second player.

Consider all losing configurations (x[i],y[i],z[i]) where x[i] ≤ y[i] ≤
z[i] ≤ 100. We can verify that Σ(x[i]+y[i]+z[i]) = 173895 for these.

Find Σ(x[i]+y[i]+z[i]) where (x[i],y[i],z[i]) ranges over the losing
configurations with x[i] ≤ y[i] ≤ z[i] ≤ 1000.


Problem 261
===========

Let us call a positive integer k a square-pivot, if there is a pair of
integers m > 0 and n ≥ k, such that the sum of the (m+1) consecutive
squares up to k equals the sum of the m consecutive squares from (n+1) on:

             (k-m)^2 + ... + k^2 = (n+1)^2 + ... + (n+m)^2.

Some small square-pivots are

 • 4: 3^2 + 4^2 = 5^2
 • 21: 20^2 + 21^2 = 29^2
 • 24: 21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2
 • 110: 108^2 + 109^2 + 110^2 = 133^2 + 134^2

Find the sum of all distinct square-pivots ≤ 10^10.


Problem 262
===========

The following equation represents the continuous topography of a
mountainous region, giving the elevation h at any point (x,y):

[Image: 262_formula1.gif]

A mosquito intends to fly from A(200,200) to B(1400,1400), without leaving
the area given by 0 ≤ x, y ≤ 1600.

Because of the intervening mountains, it first rises straight up to a
point A', having elevation f. Then, while remaining at the same elevation
f, it flies around any obstacles until it arrives at a point B' directly
above B.

First, determine f[min] which is the minimum constant elevation allowing
such a trip from A to B, while remaining in the specified area.
Then, find the length of the shortest path between A' and B', while flying
at that constant elevation f[min].

Give that length as your answer, rounded to three decimal places.

Note: For convenience, the elevation function in the image above can be
written as the following in Python (the math module must be imported):

def elevation(x, y):
    first = 5000.0 - ((x**2 + y**2 + x*y)/200.0) + (12.5*(x+y))
    second = (0.000001 * (x**2 + y**2)) - (0.0015*(x+y)) + 0.7
    return first * math.exp(-abs(second))


Problem 263
===========

Consider the number 6. The divisors of 6 are: 1,2,3 and 6.
Every number from 1 up to and including 6 can be written as a sum of
distinct divisors of 6: 1=1, 2=2, 3=1+2, 4=1+3, 5=2+3, 6=6.
A number n is called a practical number if every number from 1 up to and
including n can be expressed as a sum of distinct divisors of n.

A pair of consecutive prime numbers with a difference of six is called a
sexy pair (since "sex" is the Latin word for "six"). The first sexy pair
is (23, 29).

We may occasionally find a triple-pair, which means three consecutive sexy
prime pairs, such that the second member of each pair is the first member
of the next pair.

We shall call a number n such that:

 • (n-9, n-3), (n-3,n+3), (n+3, n+9) form a triple-pair, and
 • the numbers n-8, n-4, n, n+4 and n+8 are all practical,

an engineers’ paradise.

Find the sum of the first four engineers’ paradises.


Problem 264
===========

Consider all the triangles having:

 • All their vertices on lattice points.
 • Circumcentre at the origin O.
 • Orthocentre at the point H(5, 0).

There are nine such triangles having a perimeter ≤ 50.
Listed and shown in ascending order of their perimeter, they are:

A(-4, 3), B(5, 0), C(4, -3)
A(4, 3),  B(5, 0), C(-4, -3)
A(-3, 4), B(5, 0), C(3, -4)

A(3, 4),  B(5, 0),  C(-3, -4)
A(0, 5),  B(5, 0),  C(0, -5)
A(1, 8),  B(8, -1), C(-4, -7)

A(8, 1),  B(1, -8), C(-4, 7)
A(2, 9),  B(9, -2), C(-6, -7)
A(9, 2),  B(2, -9), C(-6, 7)

[Image: 264_TriangleCentres.gif]

The sum of their perimeters, rounded to four decimal places, is 291.0089.

Find all such triangles with a perimeter ≤ 10^5.
Enter as your answer the sum of their perimeters rounded to four decimal
places.


Problem 265
===========

2^N binary digits can be placed in a circle so that all the N-digit
clockwise subsequences are distinct.

For N=3, two such circular arrangements are possible, ignoring rotations:

[Image: 265_BinaryCircles.gif]

For the first arrangement, the 3-digit subsequences, in clockwise order,
are: 000, 001, 010, 101, 011, 111, 110 and 100.

Each circular arrangement can be encoded as a number by concatenating the
binary digits starting with the subsequence of all zeros as the most
significant bits and proceeding clockwise. The two arrangements for N=3
are thus represented as 23 and 29:

                           00010111 [2] = 23
                           00011101 [2] = 29

Calling S(N) the sum of the unique numeric representations, we can see
that S(3) = 23 + 29 = 52.

Find S(5).


Problem 266
===========

The divisors of 12 are: 1, 2, 3, 4, 6, and 12.
The largest divisor of 12 that does not exceed the square root of 12 is 3.
We shall call the largest divisor of an integer n that does not exceed the
square root of n the pseudo square root (PSR) of n.
It can be seen that PSR(3102)=47.

Let p be the product of the primes below 190. Find PSR(p) mod 10^16.

