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GeezerVison is showing github.com — JavaScript/Project-Euler/Problem012.js at master · TheAlgorithms/JavaScript Expand file tree/ Copy pathProblem012.jsMore file actions 64 lines (60 loc) · 1.87 KB / Copy pathProblem012.jsFile metadata and controls64 lines (60 loc) · 1.87 KB 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 /** * Problem 12 - Highly divisible triangular number * * https://projecteuler.net/problem=11 * * 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? */ /** * Gets number of divisors of a given number * @params num The number whose divisors to find */ const getNumOfDivisors = (num) => { // initialize numberOfDivisors let numberOfDivisors = 0 // if one divisor less than sqrt(num) exists // then another divisor greater than sqrt(n) exists and its value is num/i const sqrtNum = Math.sqrt(num) for (let i = 0; i <= sqrtNum; i++) { // check if i divides num if (num % i === 0) { if (i === sqrtNum) { // if both divisors are equal, i.e., num is perfect square, then only 1 divisor numberOfDivisors++ } else { // 2 divisors, one of them is less than sqrt(n), other greater than sqrt(n) numberOfDivisors += 2 } } } return numberOfDivisors } /** * Loops till first triangular number with 500 divisors is found */ const firstTriangularWith500Divisors = () => { let triangularNum // loop forever until numOfDivisors becomes greater than or equal to 500 for (let n = 1; ; n++) { // nth triangular number is (1/2)*n*(n+1) by Arithmetic Progression triangularNum = (1 / 2) * n * (n + 1) if (getNumOfDivisors(triangularNum) >= 500) return triangularNum } } export { firstTriangularWith500Divisors } |
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