freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-12-highly-divisible-triangular-number.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f3781000cf542c50fe8b	Problem 12: Highly divisible triangular number	5	301746	problem-12-highly-divisible-triangular-number

--description--

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

1: 1

3: 1, 3

6: 1, 2, 3, 6

10: 1, 2, 5, 10

15: 1, 3, 5, 15

21: 1, 3, 7, 21

28: 1, 2, 4, 7, 14, 28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over n divisors?

--hints--

divisibleTriangleNumber(5) should return a number.

assert(typeof divisibleTriangleNumber(5) === 'number');

divisibleTriangleNumber(5) should return 28.

assert.strictEqual(divisibleTriangleNumber(5), 28);

divisibleTriangleNumber(23) should return 630.

assert.strictEqual(divisibleTriangleNumber(23), 630);

divisibleTriangleNumber(167) should return 1385280.

assert.strictEqual(divisibleTriangleNumber(167), 1385280);

divisibleTriangleNumber(374) should return 17907120.

assert.strictEqual(divisibleTriangleNumber(374), 17907120);

divisibleTriangleNumber(500) should return 76576500.

assert.strictEqual(divisibleTriangleNumber(500), 76576500);

--seed--

--seed-contents--

function divisibleTriangleNumber(n) {

  return true;
}

divisibleTriangleNumber(500);

--solutions--

function divisibleTriangleNumber(n) {
  if (n === 1) return 3;
  let counter = 1;
  let triangleNumber = counter++;


 while (noOfFactors(triangleNumber) < n) {
   triangleNumber += counter++;
 }
return triangleNumber;
}

function noOfFactors(num) {
  const primeFactors = getPrimeFactors(num);
  let prod = 1;
  for(let p in primeFactors) {
    prod *= (primeFactors[p] + 1)
  }
  return prod;
}

function getPrimeFactors(num) {
  let n = num;
  let primes = {};

  let p = 2;
  let sqrt = Math.sqrt(num);

  function checkAndUpdate(inc) {
    if (n % p === 0) {
      const curr = primes[p];
      if (curr) {
        primes[p]++
      } else {
        primes[p] = 1;
      }
      n /= p;
    } else {
      p += inc;
    }
  }

  while(p === 2 && p <= n) {
    checkAndUpdate(1);
  }

  while (p <= n && p <= sqrt) {
    checkAndUpdate(2);
  }
  if(Object.keys(primes).length === 0) {
    primes[num] = 1;
  } else if(n !== 1) {
    primes[n] = 1;
  }
  return primes;
}