Project Euler - Problem 21

Just thought i'd post some thing on here since i haven't in a while. I wanted to start getting back into coding so i went to Project Euler. I randomly browsed some problems until I came across problem 21. Here is 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.

And this is the code i came up with to solve ( Notice: The code is not optimized! )

int sumimifier(int a)
{
int b = 1, count1 = 0, count2 = 0;
while(b < a)
{
if(a%b == 0)
{
count1 += b;
}
b++;
}
b = 1;
while(b < count1)
{
if(count1%b == 0)
{
count2 += b;
}
b++;
}
if(count2 == a && a != count1)
{
return(count1);
}
else
{
return 0;
}
}


int _tmain(int argc, _TCHAR* argv[])
{
int a, b = 0;

for( a = 0; a < 10000; a++ )
{
b += sumimifier( a );
}

printf( "%i", b );
getchar();
}

Thats basically it. If you have any better ways please post :)