2021-10-05 19:28:17 +02:00
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/**
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* Problem 20 - Factorial digit sum
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*
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* @see {@link https://projecteuler.net/problem=20}
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*
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* n! means n × (n − 1) × ... × 3 × 2 × 1
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*
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* For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
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* and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27
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*
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* Find the sum of the digits in the number 100!
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*/
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2020-10-17 23:36:20 +03:00
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2021-10-05 19:50:30 +02:00
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const factorialDigitSum = (n = 100) => {
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2021-10-05 19:28:17 +02:00
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// Consider each digit*10^exp separately, right-to-left ([units, tens, ...]).
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2021-10-05 19:50:30 +02:00
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const digits = [1]
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2020-10-17 23:36:20 +03:00
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2021-10-05 19:50:30 +02:00
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for (let x = 2; x <= n; x++) {
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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] * x + carry
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carry = Math.floor(prod / 10)
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digits[exp] = prod % 10
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2021-10-05 19:28:17 +02:00
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}
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while (carry > 0) {
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2021-10-05 19:50:30 +02:00
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digits.push(carry % 10)
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carry = Math.floor(carry / 10)
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2021-10-05 19:28:17 +02:00
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}
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}
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2020-10-17 23:36:20 +03:00
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2021-10-05 19:28:17 +02:00
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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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2020-10-17 23:36:20 +03:00
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}
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2021-10-08 13:26:10 +02:00
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export { factorialDigitSum }
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