64 lines
2.2 KiB
Java
64 lines
2.2 KiB
Java
/*
|
|
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?
|
|
*/
|
|
|
|
public class Problem_12_Highly_Divisible_Triangular_Number {
|
|
|
|
/* returns the nth triangle number; that is, the sum of all the natural numbers less than, or equal to, n */
|
|
public static int triangleNumber(int n) {
|
|
int sum = 0;
|
|
for (int i = 0; i <= n; i++)
|
|
sum += i;
|
|
return sum;
|
|
}
|
|
|
|
public static void main(String[] args) {
|
|
|
|
long start = System.currentTimeMillis(); // start the stopwatch
|
|
|
|
int j = 0; // j represents the jth triangle number
|
|
int n = 0; // n represents the triangle number corresponding to j
|
|
int numberOfDivisors = 0; // number of divisors for triangle number n
|
|
|
|
while (numberOfDivisors <= 500) {
|
|
|
|
// resets numberOfDivisors because it's now checking a new triangle number
|
|
// and also sets n to be the next triangle number
|
|
numberOfDivisors = 0;
|
|
j++;
|
|
n = triangleNumber(j);
|
|
|
|
// for every number from 1 to the square root of this triangle number,
|
|
// count the number of divisors
|
|
for (int i = 1; i <= Math.sqrt(n); i++)
|
|
if (n % i == 0)
|
|
numberOfDivisors++;
|
|
|
|
// 1 to the square root of the number holds exactly half of the divisors
|
|
// so multiply it by 2 to include the other corresponding half
|
|
numberOfDivisors *= 2;
|
|
}
|
|
|
|
long finish = System.currentTimeMillis(); // stop the stopwatch
|
|
|
|
System.out.println(n);
|
|
System.out.println("Time taken: " + (finish - start) + " milliseconds");
|
|
}
|
|
}
|