problem set.txt

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.


   Answer: e1edf9d1967ca96767dcc2b2d6df69f4


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, ...

   By considering the terms in the Fibonacci sequence whose values do not
   exceed four million, find the sum of the even-valued terms.


   Answer: 4194eb91842c8e7e6df099ca73c38f28


Problem 3
=========


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

   What is the largest prime factor of the number 600851475143 ?


   Answer: 94c4dd41f9dddce696557d3717d98d82


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.


   Answer: d4cfc27d16ea72a96b83d9bdef6ce2ec


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


   Answer: bc0d0a22a7a46212135ed0ba77d22f3a


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.


   Answer: 867380888952c39a131fe1d832246ecc


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 10 001st prime number?


   Answer: 8c32ab09ec0210af60d392e9b2009560


Problem 8
=========


   The four adjacent digits in the 1000-digit number that have the greatest
   product are 9 × 9 × 8 × 9 = 5832.

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

   Find the thirteen adjacent digits in the 1000-digit number that have the
   greatest product. What is the value of this product?


   Answer: 0f53ea7949d32ef24f9186207600403c


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.


   Answer: 24eaa9820350012ff678de47cb85b639


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.


   Answer: d915b2a9ac8749a6b837404815f1ae25


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 the same
   direction (up, down, left, right, or diagonally) in the 20×20 grid?


   Answer: 678f5d2e1eaa42f04fa53411b4f441ac


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


   The sequence of triangle numbers is generated by adding the natural
   numbers. So the 7^th 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?


   Answer: 8091de7d285989bbfa9a2f9f3bdcc7c0


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

   Answer: 361113f19fd302adc31268f8283a4f2d


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.


   Answer: 5052c3765262bb2c6be537abd60b305e


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


   Starting in the top left corner of a 2×2 grid, and only being able to move
   to the right and down, there are exactly 6 routes to the bottom right
   corner.

   How many such routes are there through a 20×20 grid?


   p_015.gif
   Answer: 928f3957168ac592c4215dcd04e0b678


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?


   Answer: 6a5889bb0190d0211a991f47bb19a777


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.


   Answer: 6a979d4a9cf85135408529edc8a133d0


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, [1]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)


   Visible links
   1. problem=67
   Answer: 708f3cf8100d5e71834b1db77dfa15d6


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)?


   Answer: a4a042cf4fd6bfb47701cbc8a1653ada


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


   n! means n × (n − 1) × ... × 3 × 2 × 1

   For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
   and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 =
   27.

   Find the sum of the digits in the number 100!


   Answer: 443cb001c138b2561a0d90720d6ce111


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.


   Answer: 51e04cd4e55e7e415bf24de9e1b0f3ff


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


   Using [1]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?


   Visible links
   1. names.txt
   Answer: f2c9c91cb025746f781fa4db8be3983f


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 n is called deficient if the sum of its proper divisors is less
   than n and it is called abundant if this sum exceeds n.

   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.


   Answer: 2c8258c0604152962f7787571511cf28


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?


   Answer: 7f155b45cb3f0a6e518d59ec348bff84


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?


   Answer: a376802c0811f1b9088828288eb0d3f0


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.


   Answer: 6aab1270668d8cac7cef2566a1c5f569


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


   Euler discovered the remarkable quadratic formula:

                                  n² + 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, 41² + 41 + 41 is clearly
   divisible by 41.

   The incredible formula  n² − 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:

     n² + 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.


   Answer: 69d9e3218fd7abb6ff453ea96505183d


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 the numbers on the diagonals is 101.

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


   Answer: 0d53425bd7c5bf9919df3718c8e49fa6


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?


   Answer: 6f0ca67289d79eb35d19decbc0a08453


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.


   Answer: 27a1779a8a8c323a307ac8a70bc4489d


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


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

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

   It is possible to make £2 in the following way:

     1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p

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


   Answer: 142dfe4a33d624d2b830a9257e96726d


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.

   Answer: 100f6e37d0b0564490a2ee27eff0660d


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.


   Answer: f899139df5e1059396431415e770c6dd


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.


   Answer: 60803ea798a0c0dfb7f36397d8d4d772


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?


   Answer: b53b3a3d6ab90ce0268229151c9bde11


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.)


   Answer: 0e175dc2f28833885f62e7345addff03


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.


   Answer: cace46c61b00de1b60874936a093981d


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?


   Answer: f2a29ede8dc9fae7926dc7a4357ac25e


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?


   Answer: fa83a11a198d5a7f0bf77a1987bcd006


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


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

                     0.123456789101112131415161718192021...

   It can be seen that the 12^th 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]


   Answer: 6f3ef77ac0e3619e98159e9b6febf557


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?


   Answer: d0a1bd6ab4229b2d0754be8923431404


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


   The n^th term of the sequence of triangle numbers is given by, t[n] =
   ½n(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 [1]words.txt, a 16K text file containing nearly two-thousand common
   English words, how many are triangle words?


   Visible links
   1. words.txt
   Answer: 82aa4b0af34c2313a562076992e50aa3


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 1^st digit, d[2] be the 2^nd 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.


   Answer: 115253b7721af0fdff25cd391dfc70cf


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 are pentagonal and D = |P[k] − P[j]| is minimised; what is
   the value of D?


   Answer: 2c2556cb85621309ca647465ffa62370


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.


   Answer: 30dfe3e3b286add9d12e493ca7be63fc


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?


   Answer: 89abe98de6071178edb1b28901a8f459


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 = 2² × 7 × 23
   645 = 3 × 5 × 43
   646 = 2 × 17 × 19.

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


   Answer: 748f517ecdc29106e2738f88aa7530f4


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.


   Answer: 0829124724747ae1c65da8cae5263346


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?


   Answer: 0b99933d3e2a9addccbb663d46cbb592


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?


   Answer: 73229bab6c5dc1c7cf7a4fa123caf6bc