Solution to Project Euler Problem 48: Self powers - The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317. Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.

• Java Code

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

import java.math.BigInteger;

public class SikademyEulerSolution {
	
	public String run() {
		BigInteger modulus = BigInteger.TEN.pow(10);
		BigInteger sum = BigInteger.ZERO;
		for (int i = 1; i <= 1000; i++)
			sum = sum.add(BigInteger.valueOf(i).modPow(BigInteger.valueOf(i), modulus));
		return sum.mod(modulus).toString();
	}
	
	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():
	MOD = 10**10
	ans = sum(pow(i, i, MOD) for i in range(1, 1001)) % MOD
	return str(ans)


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