Project Euler – Problem 76 Solution
Problem 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 …
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Project Euler – Problem 63 Solution
Problem The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power. How many n-digit positive integers exist which are also an nth power? Solution
Project Euler – Problem 62 Solution
Problem The cube, 41063625 (3453), can be permuted to produce two other cubes: 56623104 (3843) and 66430125 (4053). 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. Solution
Project Euler – Problem 112 Solution
Problem 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 …
Project Euler – Problem 51 Solution
Problem By replacing the 1st digit of *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, 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 among the ten generated …
Project Euler – Problem 39 Solution
Problem 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? Solution
Project Euler – Problem 31 Solution
Problem 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: 1x£1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p How …
Project Euler – Problem 70 Solution
Problem Euler’s Totient function, ?(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, ?(9)=6. The number 1 …
Project Euler – Problem 87 Solution
Problem 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 = 22 + 23 + 24 33 = 32 + 23 + 24 49 = 52 + …
Project Euler – Problem 92 Solution
Problem 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, Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting …