Solution to Project Euler Problem 45: Triangular, pentagonal, and hexagonal - Triangle, pentagonal, and hexagonal numbers are generated by the following formulae: Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ... Pentagonal Pn=n(3n−1)/2 1, 5, 12, 22, 35, ... Hexagonal Hn=n(2n−1) 1, 6, 15, 28, 45, ... It can be verified that T285 = P165 = H143 = 40755. Find the next triangle number that is also pentagonal and hexagonal.

• Java Code

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

public class SikademyEulerSolution {
	
	public String run() {
		int i = 286;
		int j = 166;
		int k = 144;
		while (true) {
			long triangle = (long)i * (i + 1) / 2;
			long pentagon = (long)j * (j * 3 - 1) / 2;
			long hexagon  = (long)k * (k * 2 - 1);
			long min = Math.min(Math.min(triangle, pentagon), hexagon);
			if (min == triangle && min == pentagon && min == hexagon)
				return Long.toString(min);
			if (min == triangle) i++;
			if (min == pentagon) j++;
			if (min == hexagon ) k++;
		}
	}
	

	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():
	i = 286
	j = 166
	k = 144
	while True:
		triangle = i * (i + 1) // 2
		pentagon = j * (j * 3 - 1) // 2
		hexagon  = k * (k * 2 - 1)
		minimum = min(triangle, pentagon, hexagon)
		if minimum == max(triangle, pentagon, hexagon):
			return str(triangle)
		if minimum == triangle: i += 1
		if minimum == pentagon: j += 1
		if minimum == hexagon : k += 1


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