* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
		
			
				
	
	
		
			140 lines
		
	
	
		
			2.9 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			140 lines
		
	
	
		
			2.9 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
---
 | 
						|
id: 5900f3781000cf542c50fe8b
 | 
						|
title: 'Problem 12: Highly divisible triangular number'
 | 
						|
challengeType: 5
 | 
						|
forumTopicId: 301746
 | 
						|
dashedName: 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:
 | 
						|
 | 
						|
<div style='text-align: center;'>1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...</div>
 | 
						|
 | 
						|
Let us list the factors of the first seven triangle numbers:
 | 
						|
 | 
						|
<div style='padding-left: 4em;'><b>1:</b> 1</div>
 | 
						|
<div style='padding-left: 4em;'><b>3:</b> 1, 3</div>
 | 
						|
<div style='padding-left: 4em;'><b>6:</b> 1, 2, 3, 6</div>
 | 
						|
<div style='padding-left: 4em;'><b>10:</b> 1, 2, 5, 10</div>
 | 
						|
<div style='padding-left: 4em;'><b>15:</b> 1, 3, 5, 15</div>
 | 
						|
<div style='padding-left: 4em;'><b>21:</b> 1, 3, 7, 21</div>
 | 
						|
<div style='padding-left: 4em;'><b>28:</b> 1, 2, 4, 7, 14, 28</div>
 | 
						|
 | 
						|
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.
 | 
						|
 | 
						|
```js
 | 
						|
assert(typeof divisibleTriangleNumber(5) === 'number');
 | 
						|
```
 | 
						|
 | 
						|
`divisibleTriangleNumber(5)` should return 28.
 | 
						|
 | 
						|
```js
 | 
						|
assert.strictEqual(divisibleTriangleNumber(5), 28);
 | 
						|
```
 | 
						|
 | 
						|
`divisibleTriangleNumber(23)` should return 630.
 | 
						|
 | 
						|
```js
 | 
						|
assert.strictEqual(divisibleTriangleNumber(23), 630);
 | 
						|
```
 | 
						|
 | 
						|
`divisibleTriangleNumber(167)` should return 1385280.
 | 
						|
 | 
						|
```js
 | 
						|
assert.strictEqual(divisibleTriangleNumber(167), 1385280);
 | 
						|
```
 | 
						|
 | 
						|
`divisibleTriangleNumber(374)` should return 17907120.
 | 
						|
 | 
						|
```js
 | 
						|
assert.strictEqual(divisibleTriangleNumber(374), 17907120);
 | 
						|
```
 | 
						|
 | 
						|
`divisibleTriangleNumber(500)` should return 76576500.
 | 
						|
 | 
						|
```js
 | 
						|
assert.strictEqual(divisibleTriangleNumber(500), 76576500);
 | 
						|
```
 | 
						|
 | 
						|
# --seed--
 | 
						|
 | 
						|
## --seed-contents--
 | 
						|
 | 
						|
```js
 | 
						|
function divisibleTriangleNumber(n) {
 | 
						|
 | 
						|
  return true;
 | 
						|
}
 | 
						|
 | 
						|
divisibleTriangleNumber(500);
 | 
						|
```
 | 
						|
 | 
						|
# --solutions--
 | 
						|
 | 
						|
```js
 | 
						|
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;
 | 
						|
}
 | 
						|
```
 |