2.0 KiB
2.0 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f39e1000cf542c50feb1 | Problem 50: Consecutive prime sum | 5 | 302161 | 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.
assert(typeof consecutivePrimeSum(1000) === 'number');
consecutivePrimeSum(1000)
should return 953.
assert.strictEqual(consecutivePrimeSum(1000), 953);
consecutivePrimeSum(1000000)
should return 997651.
assert.strictEqual(consecutivePrimeSum(1000000), 997651);
--seed--
--seed-contents--
function consecutivePrimeSum(limit) {
return true;
}
consecutivePrimeSum(1000000);
--solutions--
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;
}
}
}