# Project Euler

## 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 …

## Project Euler – Problem 50 Solution

Problem 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, …

## Project Euler – Problem 57 Solution

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

## Project Euler – Problem 58 Solution

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

## Project Euler – Problem 72 Solution

Problem 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, …

## Project Euler – Problem 69 Solution

Problem Euler’s Totient function, ?(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, ?(9)=6. It can be seen that n=6 produces …

## Project Euler – Problem 73 Solution

Problem 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, …