--- id: 5900f39e1000cf542c50feb1 title: 'Problem 50: Consecutive prime sum' challengeType: 5 forumTopicId: 302161 dashedName: problem-50-consecutive-prime-sum --- # --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? # --hints-- `consecutivePrimeSum(1000)` should return a number. ```js assert(typeof consecutivePrimeSum(1000) === 'number'); ``` `consecutivePrimeSum(1000)` should return 953. ```js assert.strictEqual(consecutivePrimeSum(1000), 953); ``` `consecutivePrimeSum(1000000)` should return 997651. ```js assert.strictEqual(consecutivePrimeSum(1000000), 997651); ``` # --seed-- ## --seed-contents-- ```js function consecutivePrimeSum(limit) { return true; } consecutivePrimeSum(1000000); ``` # --solutions-- ```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; } } } ```