Solution to Problem 21 on Project Euler


The problem:

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.

My solution:

#include <stdio.h>

int main(){
int i,j,z,sumA,sumB,sumPairs;

sumPairs = 0;

for (i=1;i<10000;i++){
  sumA = 0;
  for (j=1;j<i;j++){
    if (i%j==0)    
      sumA += j;
  }

  sumB = 0;
  for (z=1;z<sumA;z++){
    if (sumA%z==0)
      sumB += z;
  }

  if (sumB == i && sumB != sumA)
    sumPairs += i;  
}

printf("%dn",sumPairs);

return 0;
}

One thought on “Solution to Problem 21 on Project Euler

  1. Yavuz Asmalı

    Thank you for the solution 21. I was looking for this and your solution is the cleraest one among all the others .. 🙂

    Reply

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