Problem 21
Statement
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
The code, in C.
#include <stdio.h>
int factSum(int a)
{
int i,b=a,s=1;
for(i=2;i<b;i++){
if(!(a%i)){
s+=i;
s+=a/i;
b=a/i;
}
}
return s;
}
int amiSum(int max)
{
int i,f,s=0;
for(i=0;i<max;i++){
f=factSum(i);
if(i==factSum(f) && f<max && i!=f)
s+=i;
}
return s;
}
int main(void)
{
printf("%d\n",amiSum(10000));
return 0;
}
Answer
31626
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