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**Problem:**

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?

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**Solution:**

997651

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**Code:**

The solution may include methods that will be found here: Library.java .

The solution may include methods that will be found here: Library.java .

public interface EulerSolution{ public String run(); }

/* * Solution to Project Euler problem 50 * By Nayuki Minase * * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */ public final class p050 implements EulerSolution { public static void main(String[] args) { System.out.println(new p050().run()); } private static final int LIMIT = Library.pow(10, 6); public String run() { boolean[] isPrime = Library.listPrimality(LIMIT); int[] primes = Library.listPrimes(LIMIT); long maxSum = 0; int maxRun = -1; for (int i = 0; i < primes.length; i++) { // For each index of a starting prime number int sum = 0; for (int j = i; j < primes.length; j++) { // For each end index (inclusive) sum += primes[j]; if (sum > LIMIT) break; else if (j - i > maxRun && sum > maxSum && isPrime[sum]) { maxSum = sum; maxRun = j - i; } } } return Long.toString(maxSum); } }

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