Solution to Project Euler Problem 32: Pandigital products - 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.

• Java Code

package sikademy;
/**
 *
 * @author Archangel Macsika
 * Copyright (c) Sikademy. All rights reserved
 */

import java.util.Arrays;

public class SikademyEulerSolution {
	
	public String run() {
		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");
	}

	public static void main(String[] args) {
                SikademyEulerSolution solution = new SikademyEulerSolution();
		System.out.println(solution.run());
	}	
}

• Python Code

# 
# @author Archangel Macsika
# Copyright (c) Sikademy. All rights reserved.
# 

def compute():

	ans = sum(i for i in range(1, 10000) if has_pandigital_product(i))
	return str(ans)


def has_pandigital_product(n):
	# Find and examine all factors of n
	for i in range(1, sqrt(n) + 1):
		if n % i == 0:
			temp = str(n) + str(i) + str(n // i)
			if "".join(sorted(temp)) == "123456789":
				return True
	return False

def sqrt(x):
	assert x >= 0
	i = 1
	while i * i <= x:
		i *= 2
	y = 0
	while i > 0:
		if (y + i)**2 <= x:
			y += i
		i //= 2
	return y


if __name__ == "__main__":
	print(compute())