## Problem:

A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1, a2, ... , ak} is called a product-sum number: N = a1 + a2 + ... + ak = a1 [×] a2 [×] ... [×] ak.

For example, 6 = 1 + 2 + 3 = 1 [×] 2 [×] 3.

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k = 2, 3, 4, 5, and 6 are as follows.

k=2: 4 = 2 [×] 2 = 2 + 2
k=3: 6 = 1 [×] 2 [×] 3 = 1 + 2 + 3
k=4: 8 = 1 [×] 1 [×] 2 [×] 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 [×] 1 [×] 2 [×] 2 [×] 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 [×] 1 [×] 1 [×] 1 [×] 2 [×] 6 = 1 + 1 + 1 + 1 + 2 + 6

Hence for 2[≤]k[≤]6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2[≤]k[≤]12 is {4, 6, 8, 12, 15, 16}, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2[≤]k[≤]12000?

932718654

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

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 38 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */import java.util.Arrays;public final class p038 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p038().run());	}			public String run() {		int max = -1;		for (int n = 2; n <= 9; n++) {			for (int i = 1; i < Library.pow(10, 9 / n); i++) {				String concat = "";				for (int j = 1; j <= n; j++)					concat += i * j;				if (isPandigital(concat))					max = Math.max(Integer.parseInt(concat), max);			}		}		return Integer.toString(max);	}			private static boolean isPandigital(String s) {		if (s.length() != 9)			return false;		char[] temp = s.toCharArray();		Arrays.sort(temp);		return new String(temp).equals("123456789");	}	}

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