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chore: Merge pull request TheAlgorithms#745 from lvlte/ProjectEuler/016
Project Euler - Problem 16
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DIRECTORY.md

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* [Problem013](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem013.js)
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* [Problem014](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem014.js)
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* [Problem015](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem015.js)
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* [Problem016](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem016.js)
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* [Problem020](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem020.js)
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* [Problem1](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem1.js)
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* [Problem10](https://github.com/TheAlgorithms/Javascript/blob/master/Project-Euler/Problem10.js)

Project-Euler/Problem016.js

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/**
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* Problem 16 - Power digit sum
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*
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* @see {@link https://projecteuler.net/problem=16}
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*
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* 2¹⁵ = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
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*
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* What is the sum of the digits of the number 2¹⁰⁰⁰ ?
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*/
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/**
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* Returns the power digit sum of n^pow.
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*
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* @param {number} [n=2]
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* @param {number} [pow=1000]
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* @returns {number}
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*/
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const powerDigitSum = function (n = 2, pow = 1000) {
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// The idea is to consider each digit (d*10^exp) separately, right-to-left.
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// digits = [units, tens, ...]
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const digits = [n]
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let p = 1
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while (++p <= pow) {
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let carry = 0
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for (let exp = 0; exp < digits.length; exp++) {
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const prod = digits[exp] * n + carry
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carry = Math.floor(prod / 10)
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digits[exp] = prod % 10
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}
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while (carry > 0) {
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digits.push(carry % 10)
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carry = Math.floor(carry / 10)
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}
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}
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// (digits are reversed but we only want the sum so it doesn't matter)
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return digits.reduce((prev, current) => prev + current, 0)
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}
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export { powerDigitSum }

Project-Euler/test/Problem016.test.js

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import { powerDigitSum } from '../Problem016'
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describe('Check Problem 16 - Power digit sum', () => {
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it('Power digit sum of 2^15', () => {
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expect(powerDigitSum(2, 15)).toBe(26)
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})
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it('Power digit sum of 2^1000', () => {
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expect(powerDigitSum()).toBe(1366)
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expect(powerDigitSum(2, 1000)).toBe(1366)
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})
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it('Power digit sum of 3^5000', () => {
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expect(powerDigitSum(3, 5000)).toBe(11097)
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})
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})

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