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.

Solution:31626

public class Problem21 {

public static void main(String[] args) {

int sum = 0;

for (int i = 1; i < 10000; i++) {

int divSum = getProperDivisorsSum(i);

if(i!=divSum && i==getProperDivisorsSum(divSum))

{

sum+=i;

}

}

System.out.println(sum);

}

private static int getProperDivisorsSum(int num){

int sum = 0;

for (int j = 1; j < num; j++) {

if(num%j==0){

sum += j;

}

}

return sum;

}

}

public static void main(String[] args) {

int sum = 0;

for (int i = 1; i < 10000; i++) {

int divSum = getProperDivisorsSum(i);

if(i!=divSum && i==getProperDivisorsSum(divSum))

{

sum+=i;

}

}

System.out.println(sum);

}

private static int getProperDivisorsSum(int num){

int sum = 0;

for (int j = 1; j < num; j++) {

if(num%j==0){

sum += j;

}

}

return sum;

}

}