freeCodeCamp/problem-20-factorial-digit-sum.english.md at ef79ddd0de35a4e10b8b4588e67938f86d2bae10

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mrugesh 22afc2a0ca feat(learn): python certification projects (#38216 )

Co-authored-by: Oliver Eyton-Williams <ojeytonwilliams@gmail.com>
Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com>
Co-authored-by: Beau Carnes <beaucarnes@gmail.com>

2020-05-27 13:19:08 +05:30

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id, challengeType, isHidden, title, forumTopicId

id	challengeType	isHidden	title	forumTopicId
5900f3801000cf542c50fe93	5	false	Problem 20: Factorial digit sum	301839

Description

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 n!

Instructions

Tests

tests:
  - text: <code>sumFactorialDigits(10)</code> should return a number.
    testString: assert(typeof sumFactorialDigits(10) === 'number');
  - text: <code>sumFactorialDigits(10)</code> should return 27.
    testString: assert.strictEqual(sumFactorialDigits(10), 27);
  - text: <code>sumFactorialDigits(25)</code> should return 72.
    testString: assert.strictEqual(sumFactorialDigits(25), 72);
  - text: <code>sumFactorialDigits(50)</code> should return 216.
    testString: assert.strictEqual(sumFactorialDigits(50), 216);
  - text: <code>sumFactorialDigits(75)</code> should return 432.
    testString: assert.strictEqual(sumFactorialDigits(75), 432);
  - text: <code>sumFactorialDigits(100)</code> should return 648.
    testString: assert.strictEqual(sumFactorialDigits(100), 648);

Challenge Seed

function sumFactorialDigits(n) {
  // Good luck!
  return n;
}

sumFactorialDigits(100);

Solution

let factorial = (n) => n <= 1 ? BigInt(n) : BigInt(n) * BigInt(factorial(--n));

let sumDigits = n => n.toString().split('').map(x => parseInt(x)).reduce((a,b) => a + b);

function sumFactorialDigits(n) {
  return sumDigits(factorial(n));
}

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