euler

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