## 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.

45228

## Code: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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