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* chore: Switch to Node 20 + Vitest * chore: migrate to vitest mock functions * chore: code style (switch to prettier) * test: re-enable long-running test Seems the switch to Node 20 and Vitest has vastly improved the code's and / or the test's runtime! see #1193 * chore: code style * chore: fix failing tests * Updated Documentation in README.md * Update contribution guidelines to state usage of Prettier * fix: set prettier printWidth back to 80 * chore: apply updated code style automatically * fix: set prettier line endings to lf again * chore: apply updated code style automatically --------- Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com> Co-authored-by: Lars Müller <34514239+appgurueu@users.noreply.github.com>
65 lines
1.9 KiB
JavaScript
65 lines
1.9 KiB
JavaScript
/**
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* Problem 12 - Highly divisible triangular number
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*
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* https://projecteuler.net/problem=11
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*
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* The sequence of triangle numbers is generated by adding the natural numbers.
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* So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
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*
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* The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
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* Let us list the factors of the first seven triangle numbers:
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*
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* 1: 1
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* 3: 1,3
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* 6: 1,2,3,6
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* 10: 1,2,5,10
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* 15: 1,3,5,15
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* 21: 1,3,7,21
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* 28: 1,2,4,7,14,28
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*
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* We can see that 28 is the first triangle number to have over five divisors.
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*
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* What is the value of the first triangle number to have over five hundred divisors?
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*/
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/**
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* Gets number of divisors of a given number
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* @params num The number whose divisors to find
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*/
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const getNumOfDivisors = (num) => {
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// initialize numberOfDivisors
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let numberOfDivisors = 0
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// if one divisor less than sqrt(num) exists
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// then another divisor greater than sqrt(n) exists and its value is num/i
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const sqrtNum = Math.sqrt(num)
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for (let i = 0; i <= sqrtNum; i++) {
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// check if i divides num
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if (num % i === 0) {
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if (i === sqrtNum) {
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// if both divisors are equal, i.e., num is perfect square, then only 1 divisor
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numberOfDivisors++
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} else {
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// 2 divisors, one of them is less than sqrt(n), other greater than sqrt(n)
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numberOfDivisors += 2
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}
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}
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}
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return numberOfDivisors
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}
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/**
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* Loops till first triangular number with 500 divisors is found
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*/
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const firstTriangularWith500Divisors = () => {
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let triangularNum
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// loop forever until numOfDivisors becomes greater than or equal to 500
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for (let n = 1; ; n++) {
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// nth triangular number is (1/2)*n*(n+1) by Arithmetic Progression
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triangularNum = (1 / 2) * n * (n + 1)
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if (getNumOfDivisors(triangularNum) >= 500) return triangularNum
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}
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}
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export { firstTriangularWith500Divisors }
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