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Problem 56 07 November 2003 A googol (10100) is a massive number: one followed by one-hundred zeros; 100100 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, ab, where a, b 100, what is the maximum digital sum? Solution in C# using System; using System.Collections.Generic; ..

Problem 55 24 October 2003 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 nev..

Problem 53 26 September 2003 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, 5C3 = 10. In general, nCr = n! r!(nr)! ,where r n, n! = n(n1)...321, and 0! = 1. It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066. How many, not necessarily distinct, values of nCr, for 1 ..

namespace etc { static class Etc { static BigInteger combination(int n, int r) { return (factorial(n) / factorial(r) / factorial(n - r)); } static Dictionary f5Cache = new Dictionary(); static int f5(BigInteger n) { if (f5Cache.ContainsKey(n)) return f5Cache[n]; int r = 0; for (int x = (int)Double.Parse(BigInteger.Log(n, 5).ToString()); x > 0; x--) { if (n % BigInteger.Pow(5, x) == 0) { r = x; b..

Problem 3 02 November 2001 The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? Answer: 6857 Solution in Perl my $num = 600851475143; my $p; for ( $p = 2 ; ; $p++ ) { while ( $num % $p == 0 ) { $num /= $p; } last if ( $num num2 1) (loop (add1 p) num2)] [else (display p)]))) ; imperative, one loop #lang racket (define (problem3) (define num..

Problem 2 19 October 2001 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: 4613732 Solution in Perl use List::Util qw(sum); sub ..

Problem 1 05 October 2001 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: 233168 Solution in C# using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Numerics; namespace Euler { static class Program { static..

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