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**Problem:**

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 [×] 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

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**Solution:**

45228

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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 32 * 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 p032 implements EulerSolution { public static void main(String[] args) { System.out.println(new p032().run()); } public String run() { // A candidate product has at most 4 digits. This is because if it has 5 digits, // then the two multiplicands must have at least 5 digits put together. int sum = 0; for (int i = 1; i < 10000; i++) { if (hasPandigitalProduct(i)) sum += i; } return Integer.toString(sum); } private static boolean hasPandigitalProduct(int n) { // Find and examine all factors of n for (int i = 1; i <= n; i++) { if (n % i == 0 && isPandigital("" + n + i + n/i)) return true; } return false; } 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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