101 lines
2.2 KiB
Markdown
101 lines
2.2 KiB
Markdown
---
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id: 5900f39e1000cf542c50feb1
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challengeType: 5
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title: 'Problem 50: Consecutive prime sum'
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---
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## Description
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<section id='description'>
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The prime 41, can be written as the sum of six consecutive primes:
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41 = 2 + 3 + 5 + 7 + 11 + 13
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This is the longest sum of consecutive primes that adds to a prime below one-hundred.
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The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
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Which prime, below one-million, can be written as the sum of the most consecutive primes?
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</section>
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## Instructions
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<section id='instructions'>
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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- text: <code>consecutivePrimeSum(1000)</code> should return 953.
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testString: assert.strictEqual(consecutivePrimeSum(1000), 953, '<code>consecutivePrimeSum(1000)</code> should return 953.');
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- text: <code>consecutivePrimeSum(1000000)</code> should return 997651.
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testString: assert.strictEqual(consecutivePrimeSum(1000000), 997651, '<code>consecutivePrimeSum(1000000)</code> should return 997651.');
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function consecutivePrimeSum(limit) {
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// Good luck!
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return true;
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}
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consecutivePrimeSum(1000000);
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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function consecutivePrimeSum(limit) {
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function isPrime(num) {
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if (num < 2) {
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return false;
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} else if (num === 2) {
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return true;
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}
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const sqrtOfNum = Math.floor(num ** 0.5);
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for (let i = 2; i <= sqrtOfNum + 1; i++) {
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if (num % i === 0) {
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return false;
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}
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}
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return true;
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}
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function getPrimes(limit) {
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const primes = [];
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for (let i = 0; i <= limit; i++) {
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if (isPrime(i)) primes.push(i);
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}
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return primes;
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}
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const primes = getPrimes(limit);
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let primeSum = [...primes];
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primeSum.reduce((acc, n, i) => {
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primeSum[i] += acc;
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return acc += n;
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}, 0);
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for (let j = primeSum.length - 1; j >= 0; j--) {
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for (let i = 0; i < j; i++) {
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const sum = primeSum[j] - primeSum[i];
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if (sum > limit) break;
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if (isPrime(sum) && primes.indexOf(sum) > -1) return sum;
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}
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}
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}
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```
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</section>
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