ProEuler_P12

package problems;


/*
Problem 12
08 March 2002

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

     1: 1
     3: 1,3
     6: 1,2,3,6
    10: 1,2,5,10
    15: 1,3,5,15
    21: 1,3,7,21
    28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

Answer:

FIVE HUNDRED DIVISORS?

This means the number has to be made up of at least 500 numbers, and probably a lot more.


Generate numbers until you find one with 500 divisors?...

0.5 x n(n+1)

*/

public class Problem_12 {

    public static void main(String[] args) {

        int number = 0;
        int i = 1;

        while(numberOfDivisors(number) < 500){
            number += i;
            i++;
        }

        System.out.println("The answer is: " +number +" with " + numberOfDivisors(number) +" factors.");

    } //End of main.

    private static int numberOfDivisors(int inputNumber){

        int nod = 0; // Counts Number Of Divsions
        int sqrt = (int) Math.sqrt(inputNumber);

        /* No need to check higher than the sqrt of the current number,
        because if a value higher than the square root were a factor, it
        would already have been checked. A factor of a number will always
        have a "sibling" so to speak. */

        for(int i = 1; i<= sqrt; i++){

            if(inputNumber % i == 0){
                nod += 2;

                /* You've found one factor, that means there must be
                 * another number which is a factor. So you increment
                 * by two.
                 *
                 * ALL numbers have at least two factors (except 1),
                 * 1 and the number itself.
                 */
            }

        }


        /* Correction if the number is a perfect square.
         * */

        if (sqrt * sqrt == inputNumber) {
            nod--;

            /* In other words, if the factor is the sqrt
             * of the number it has no "sibling number" it
             * is forever alone.
             *
             * Example = 9 has the factors 1, 3 and 9.
             */
        }

        return nod;


    }

}