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Problem020.js
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Problem020.js
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/**
* Problem 20 - Factorial digit sum
*
* @see {@link https://projecteuler.net/problem=20}
*
* n! means n × (n − 1) × ... × 3 × 2 × 1
*
* For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
* and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27
*
* Find the sum of the digits in the number 100!
*/
const factorialDigitSum = (n = 100) => {
// Consider each digit*10^exp separately, right-to-left ([units, tens, ...]).
const digits = [1]
for (let x = 2; x <= n; x++) {
let carry = 0
for (let exp = 0; exp < digits.length; exp++) {
const prod = digits[exp] * x + carry
carry = Math.floor(prod / 10)
digits[exp] = prod % 10
}
while (carry > 0) {
digits.push(carry % 10)
carry = Math.floor(carry / 10)
}
}
// (digits are reversed but we only want the sum so it doesn't matter)
return digits.reduce((prev, current) => prev + current, 0)
}
export { factorialDigitSum }