M such that N mod M? N/2.

Im not an expert here I dont understand the question either but
this is my understanding of the statement above..

this will display the numbers from 0 to Mth value where N is the lucky number and M is the number od iterations
then the M modulo N should be equal to the value of (N/2)

the program below will result to

the M = 7
the M = 22
the M = 37
the M = 52
the M = 67
the M = 82
the M = 97
Number of Hits: 7

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#include <stdio.h>

int M_limit=100;    //the Mth limit
int N = 15;         //your lucky number N
int H_count=0;      //number positive integers M 

void the_M(){
    
    for(int a=1; a<M_limit; a++){
            
         int M=a;
        if( (N%M) ==(N/2)){             //if we reverse M and N in the modulo the program will crash i dont know
            printf("the M = %d\n",a);   //print the M value based from the formula above i think
            H_count++;                  //
        }
    }
    printf("\nNumber of Hits: %d\n",H_count);  //print the number of integers of Ms
}

int main(){

    the_M();
    return 0;
}