Thread: need to solve simple c problem

Problem Name: Y = MX
Everyone know that the equation of a straight line is Y = MX + C. But if C = 0, then the straight line is passing through the origin. Here I always put C = 0. So our equation will be always Y = MX. Straight line is a geometrical term but my problem is not geometrical. I just use the equation Y = MX. Since Y = MX, so Y is must be divisible by M and X, and for a specific value of Y and various value of M and X, the equation must be satisfy.
But in this problem we find (N) the number of M and X that satisfy the equation with condition M must be less than X (M16 = 1 * 16 (1<16) so it is countable.
16 = 2 * 8 (2<8) so it is countable.
16 = 4 * 4 (4 = 4) so it is not countable.
16 = 8 * 2 (8>2) so it is not countable.
16 = 16 * 1 (16>1) so it is not countable.
So our answer N=2.

So your duty is to find out how many way to find M and X (M
Input: You have to provide an integer Y (1<=Y<=10000).

Output: You just print the value of N.

Sample input:
16

Sample output:
2

what output i'll get for following those input 128,500,666,1000,10000?
i need code and solution

i tried many time but can't get logic. how i improve my logic?
help to clera this program logic concept. ... Shift+R improves the quality of this image. Shift+A improves the quality of all images on this page.

Code:

#include<stdio.h>
int main()
{
    
    int a,b,c=1,i;
    scanf("%d",&a);
    for(i=2;i<=a;i+=2)
    {
        if((a%i)==0)
        {
        printf("%d ",i);
        b=(a%i);
        }

        if(b>i)
         c++;

    }

    printf("\n%d",c);
}