Project Euler > Problem 69 > Totient maximum (Java Solution)

Problem:

Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.

n Relatively Prime φ(n) n/φ(n)
2 1 1 2
3 1,2 2 1.5
4 1,3 2 2
5 1,2,3,4 4 1.25
6 1,5 2 3
7 1,2,3,4,5,6 6 1.1666...
8 1,3,5,7 4 2
9 1,2,4,5,7,8 6 1.5
10 1,3,7,9 4 2.5

It can be seen that n=6 produces a maximum n/φ(n) for n [≤] 10.

Find the value of n [≤] 1,000,000 for which n/φ(n) is a maximum.

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;
	}
	
}