## Project Euler: Problem 21 – Amicable numbers

#### Problem 21:

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).

If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

#### Idea:

Find the proper divisors of a given number and add them up. Then repeat for the sum and add up those proper divisors and see if the two numbers form a cycle, and are not the same number.

```int answer = 0;

int[] values = new int;
for(int i = 0; i < values.length; i++) {
int sum = 1; //all will have minimum value of 1
for(int divisor = 2; divisor <= Math.sqrt(i); divisor++) {
int quotient = i/divisor;
if(quotient * divisor == i) {
if(divisor != quotient) {
sum+=divisor;
sum+=quotient;
} else {
sum+=divisor;
}
}
}
values[i] = sum;
}

for(int i = 1; i < values.length; i++) {
//Sample inputs to distinguish between good and bad
//     i = 6   220
// first = 6   284
//second = 6   220