CountFactors is a problem where one must compute the factors of a given number. Note that before attempting this problem, you should read the material on prime and composite numbers .
We incremente the count by two because based on one divisor, we can find the symmetric divisor. In other words, if a is a divisor of n then n/a is also one. The only case when that does not happen is when is if the number is in the form k^2, meaning that the symmetric divisor of k is also k.
int i = 1;
int result = 0;
while (i < Math.sqrt(n)) {
if (n % i == 0) {
result += 2;
}
i++;
}
if (Math.pow(i, 2) == n) {
result += 1;
}
return result;The full solution can be found here.