Project Euler > Problem 53 > Combinatoric selections (Java Solution)

Problem:

There are exactly ten ways of selecting three from five, 12345:

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, 5C3 = 10.

In general,

nCr =
n!
r!(n[−]r)!
,where r [≤] n, n! = n[×](n[−]1)[×]...[×]3[×]2[×]1, and 0! = 1.

It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.

How many, not necessarily distinct, values of nCr, for 1 [≤] n [≤] 100, are greater than one-million?

Solution:

6857

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

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 3
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p003 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p003().run());
	}
	
	
	/* 
	 * Algorithm: Divide out all the smallest prime factors except the last one.
	 * For example, 1596 = 2 * 2 * 3 * 7 * 19. The algorithm ensures that the smallest factors will be found first.
	 * After dividing out the smallest factors, the last factor to be found will be equal to the quotient, so it must be the largest prime factor.
	 */
	public String run() {
		long n = 600851475143L;
		while (true) {
			long p = smallestFactor(n);
			if (p < n)
				n /= p;
			else
				return Long.toString(n);
		}
	}
	
	
	private static long smallestFactor(long n) {
		for (long i = 2, end = Library.sqrt(n); i <= end; i++) {
			if (n % i == 0)
				return i;
		}
		return n;  // Prime
	}
	
}