C - Find all amicable numbers between limits - c

Question

First a definition: An amicable pair of numbers consists of two different integers where the
sum of the divisors of the first integer is equal to the second integer, and the
sum of the divisors of the second integer is equal to the first integer. A perfect number is a number that equals the sum of its own divisors.
What I want to do is make a program that asks the user for a lower limit, and an upper limit and then presents him/her with all the amicable pairs (one per line) between those two limits. If there's a perfect number only one number needs to be printed (not a pair in its case).
The whole idea is pretty confusing to me, so I'm looking for some assistance.
Here's what I have to start with, I know that sumDivisors() should be more or less correct, but main() is merely checking if the two inputted numbers are amicable or not - might have to do a complete overhaul of this since I want all the pairs between two given limits.
long sumDivisors(long a)
{
long s=0,i;
for(i=1;i<=a/2;i++)
{
if(a%i==0)
{
s+=i;
}
}
return s;
}
int main()
{
long t,n,s1;
scanf("%ld",&t);
while(t--)
{
scanf("%ld",&n);
s1=sumDivisors(n);
if(n==sumDivisors(s1))
{
printf("Yes\n");
}
else printf("No\n");
}
return 0;
}

Answer 1

You could write main() like this:
int main ()
{
// assumes 1 <= min <= max
long min = 1;
long max = 10000;
for (long a = min; a <= max; a++) {
long b = sum_of_proper_divisors (a);
if (a == b) {
printf ("perfect number:\t%ld\n", a);
}
if ((a < b) && (b <= max) && (sum_of_proper_divisors (b) == a)) {
printf ("amicable pair:\t(%ld, %ld)\n", a, b);
}
}
return 0;
}

Answer 2

The most easiest and understandable way of finding amicable pairs between a range is as follows:
find amicable pairs between 1 to 2000.if you want between 1 to 3000 , just bring changes in the checking condition of for loops( i and j <= 3000).
You can give whatever range you want (by changing the initialization and checking conditions of the loops(outer loop and inner loop) .
#include<stdio.h>
int main(){
int i,j;
//outer loop.
for (i=1;i<=2000;i++){
int d1=1;
int sum1=0;
while(d1<i){
if(i%d1==0){
sum1+=d1; //sum of divisors of i
d1++;
}else
d1++;
}
//inner loop
for(j=i+1;j<=2000;j++){
int d2=1;
int sum2=0;
while(d2<j){
if(j%d2==0){
sum2+=d2;//sum of divisors of j
d2++;
}else
d2++;
}
if(sum1==j && sum2==i)
//printing amicalbe pair.
printf("(%d , %d) \n",i,j);
}
}
return 0;
}

Answer 3

Most of you might face problem understanding what amicable pairs are, let me explain it through an example 220 & 284 are said to be amicable pairs because if we find the proper divisors of 220 we get (1, 2, 4, 5, 10, 11, 20, 22, 44, 55 & 110) summing up all of them we get 284. Now, the proper divisors of 284 are (1, 2, 4, 71 & 142) summing up all of them we get 220. Similarly, the sum of the divisors of 1184 is equal 1210 & the sum of the divisors of 1210 is equal to 1184. Now, we write a program in C to find all the amicable pairs within the range of 10000.
int main()
{
int n,k;
int i=1,s1=0,s2=0;
for(k=1;k<=10000;k++)
{
n=k;
while(i<n)
{
if(n%i==0)
s1=s1+i;
i++;
}
i=1;
if(s1==n)
continue;
while(i<s1)
{
if(s1%i==0)
s2=s2+i;
i++;
}
if(n==s2)
printf("%d \n",n);
s1=0;
s2=0;
}
}