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id	title	challengeType	forumTopicId	dashedName
5900f3811000cf542c50fe94	Problem 21: Amicable numbers	5	301851	problem-21-amicable-numbers

--description--

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).

If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under n.

--hints--

sumAmicableNum(1000) should return a number.

assert(typeof sumAmicableNum(1000) === 'number');

sumAmicableNum(1000) should return 504.

assert.strictEqual(sumAmicableNum(1000), 504);

sumAmicableNum(2000) should return 2898.

assert.strictEqual(sumAmicableNum(2000), 2898);

sumAmicableNum(5000) should return 8442.

assert.strictEqual(sumAmicableNum(5000), 8442);

sumAmicableNum(10000) should return 31626.

assert.strictEqual(sumAmicableNum(10000), 31626);

--seed--

--seed-contents--

function sumAmicableNum(n) {

  return n;
}

sumAmicableNum(10000);

--solutions--

const sumAmicableNum = (n) => {
  const fsum = (n) => {
    let sum = 1;
    for (let i = 2; i <= Math.floor(Math.sqrt(n)); i++)
      if (Math.floor(n % i) === 0)
        sum += i + Math.floor(n / i);
    return sum;
  };
  let d = [];
  let amicableSum = 0;
  for (let i=2; i<n; i++) d[i] = fsum(i);
  for (let i=2; i<n; i++) {
    let dsum = d[i];
    if (d[dsum]===i && i!==dsum) amicableSum += i+dsum;
  }
  return amicableSum/2;
};

1.7 KiB Raw Blame History