Project Euler > Problem 119 > Digit power sum (Java Solution)

Problem:

The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 83 = 512. Another example of a number with this property is 614656 = 284.

We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.

You are given that a2 = 512 and a10 = 614656.

Find a30.

Solution:

171

Code:
The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 19
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p019 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p019().run());
	}
	
	
	public String run() {
		int count = 0;
		for (int y = 1901; y <= 2000; y++) {
			for (int m = 1; m <= 12; m++) {
				if (dayOfWeek(y, m, 1) == 0)  // Sunday
					count++;
			}
		}
		return Integer.toString(count);
	}
	
	
	private static int dayOfWeek(int year, int month, int day) {
		long m = mod((long)month - 3, 4800);
		long y = mod(year + m / 12, 400);
		m %= 12;
		return (int)((y + y/4 - y/100 + (13 * m + 2) / 5 + day + 2) % 7);
	}
	
	
	private static long mod(long x, long y) {
		x %= y;
		if (y > 0 && x < 0 || y < 0 && x > 0)
			x += y;
		return x;
	}
	
}