freeCodeCamp/curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.english.md

---
id: 5900f39e1000cf542c50feb1
challengeType: 5
title: 'Problem 50: Consecutive prime sum'
---

## Description
<section id='description'>
The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?
</section>

## Instructions
<section id='instructions'>

</section>

## Tests
<section id='tests'>

```yml
tests:
  - text: <code>consecutivePrimeSum(1000)</code> should return 953.
    testString: assert.strictEqual(consecutivePrimeSum(1000), 953, '<code>consecutivePrimeSum(1000)</code> should return 953.');
  - text: <code>consecutivePrimeSum(1000000)</code> should return 997651.
    testString: assert.strictEqual(consecutivePrimeSum(1000000), 997651, '<code>consecutivePrimeSum(1000000)</code> should return 997651.');

```

</section>

## Challenge Seed
<section id='challengeSeed'>

<div id='js-seed'>

```js
function consecutivePrimeSum(limit) {
  // Good luck!
  return true;
}

consecutivePrimeSum(1000000);
```

</div>



</section>

## Solution
<section id='solution'>


```js
function consecutivePrimeSum(limit) {
  function isPrime(num) {
    if (num < 2) {
      return false;
    } else if (num === 2) {
      return true;
    }
    const sqrtOfNum = Math.floor(num ** 0.5);
    for (let i = 2; i <= sqrtOfNum + 1; i++) {
      if (num % i === 0) {
        return false;
      }
    }
    return true;
  }
  function getPrimes(limit) {
    const primes = [];
    for (let i = 0; i <= limit; i++) {
      if (isPrime(i)) primes.push(i);
    }
    return primes;
  }

  const primes = getPrimes(limit);
  let primeSum = [...primes];
  primeSum.reduce((acc, n, i) => {
    primeSum[i] += acc;
    return acc += n;
  }, 0);

  for (let j = primeSum.length - 1; j >= 0; j--) {
    for (let i = 0; i < j; i++) {
      const sum = primeSum[j] - primeSum[i];
      if (sum > limit) break;
      if (isPrime(sum) && primes.indexOf(sum) > -1) return sum;
    }
  }
}
```

</section>