Solution to Project Euler Problem 33: Digit cancelling fractions - The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s. We shall consider fractions like, 30/50 = 3/5, to be trivial examples. There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator. If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

• Java Code

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

public class SikademyEulerSolution {
	
	public String run() {
		int numer = 1;
		int denom = 1;
		for (int d = 10; d < 100; d++) {
			for (int n = 10; n < d; n++) {
				int n0 = n % 10, n1 = n / 10;
				int d0 = d % 10, d1 = d / 10;
				if (n1 == d0 && n0 * d == n * d1 || n0 == d1 && n1 * d == n * d0) {
					numer *= n;
					denom *= d;
				}
			}
		}
		return Integer.toString(denom / gcd(numer, denom));
	}

        public static int gcd(int x, int y) {
		if (x < 0 || y < 0)
			throw new IllegalArgumentException("Negative number");
		while (y != 0) {
			int z = x % y;
			x = y;
			y = z;
		}
		return x;
	}

	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.
# 

import fractions

def compute():

	numer = 1
	denom = 1
	for d in range(10, 100):
		for n in range(10, d):
			n0 = n % 10
			n1 = n // 10
			d0 = d % 10
			d1 = d // 10
			if (n1 == d0 and n0 * d == n * d1) or (n0 == d1 and n1 * d == n * d0):
				numer *= n
				denom *= d
	return str(denom // fractions.gcd(numer, denom))

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