3.1 KiB
3.1 KiB
id | challengeType | title | forumTopicId |
---|---|---|---|
5900f3781000cf542c50fe8b | 5 | Problem 12: Highly divisible triangular number | 301746 |
Description
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?
Instructions
Tests
tests:
- text: <code>divisibleTriangleNumber(5)</code> should return 28.
testString: assert.strictEqual(divisibleTriangleNumber(5), 28);
- text: <code>divisibleTriangleNumber(23)</code> should return 630.
testString: assert.strictEqual(divisibleTriangleNumber(23), 630);
- text: <code>divisibleTriangleNumber(167)</code> should return 1385280.
testString: assert.strictEqual(divisibleTriangleNumber(167), 1385280);
- text: <code>divisibleTriangleNumber(374)</code> should return 17907120.
testString: assert.strictEqual(divisibleTriangleNumber(374), 17907120);
- text: <code>divisibleTriangleNumber(500)</code> should return 76576500.
testString: assert.strictEqual(divisibleTriangleNumber(500), 76576500);
Challenge Seed
function divisibleTriangleNumber(n) {
// Good luck!
return true;
}
divisibleTriangleNumber(500);
Solution
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;
}