## Problem:

Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn[−]1)n + (pn+1)n is divided by pn2.

For example, when n = 3, p3 = 5, and 43 + 63 = 280 [≡] 5 mod 25.

The least value of n for which the remainder first exceeds 109 is 7037.

Find the least value of n for which the remainder first exceeds 1010.

4179871

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

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 23 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */public final class p023 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p023().run());	}			private static final int LIMIT = 28123;		private boolean[] isAbundant = new boolean[LIMIT + 1];		public String run() {		// Compute look-up table		for (int i = 1; i < isAbundant.length; i++)			isAbundant[i] = isAbundant(i);				int sum = 0;		for (int i = 1; i <= LIMIT; i++) {			if (!isSumOf2Abundants(i))				sum += i;		}		return Integer.toString(sum);	}			private boolean isSumOf2Abundants(int n) {		for (int i = 0; i <= n; i++) {			if (isAbundant[i] && isAbundant[n - i])				return true;		}		return false;	}			private static boolean isAbundant(int n) {		if (n < 1)			throw new IllegalArgumentException();				int sum = 1;  // Sum of factors less than n		int end = Library.sqrt(n);		for (int i = 2; i <= end; i++) {			if (n % i == 0)				sum += i + n / i;		}		if (end * end == n)			sum -= end;		return sum > n;	}	}

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