need to solve simple c problem

[raihan004]

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.

#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);
}

[root4]

i need code and solution

Are you joking? please read the forum FAQ.

[raihan004]

Are you joking? please read the forum FAQ.

sorry am new. can't i ask any problem solution code ? what is my fault?

[root4]

can't i ask any problem solution code ?

Indeed. You can get some help, here, with programming issues, but nobody is going to solve any assignment.
Work on your problem first, show us some code and possible problems you face, then we'll help.

[Adak]

I have been feverishly looking for a silver platter with a gold rim - something suitable for the delivery of this code and solution.

All in vain.

[Salem]

You can only get so far by looking at code.
The real learning starts when you actually try and write some yourself.

Make an actual effort, and then we'll help you.

[raihan004]

now i inlcude my tried attetmp code . now make me my code fullfill and make clear concept

[root4]

1. Your assignment gives you explicit names for the objects so use them in your code, that gives a readability bonus (e.g. what you named "a" is "y").
2. Indent correctly your code.

You have all the pieces, but not composed correctly: to increment your counter (c), Y has to be divisible by your current M (Y%M==0) and that M must be less than (Y/M) so you have to merge your two conditionals.
I don't know why you increment M by 2.

[iMalc]

The problem comes from asking the equivalent of "do it for me" when you should be asking the equivalent of "teach me".

[raihan004]

i solve this problem . if any wrong about my logic please repaly or notify me thanks all

  

#include<stdio.h>
int main()
{

    int y,m,n=0,i,x=0;
    scanf("%d",&y);
    for(i=1;i<=y;i++)
    {
        if((y%i)==0)
        {


x=y/i;

        if(x>i)
        {
            printf("%4d",i);
         n++;
        }
        }
    }

    printf("\ntotal %d step need to do this",n);
}

:redface:

[root4]

Looks ok.
Indent your code properly.
You can merge the two conditionals with a logical 'and' operator (&&).
Your variable 'm' is unused so far, simply replace 'i' with 'm'.