n and n+1 are integers with the same number of positive divisions. Find the integers n from 1

Programming instructions:

• Include the necessary header file.
• In the main function, declare the required variables.
• Use for-loop to call the divisorcount() function for each number.
• Call the divisorcount() function for a number and store the value returned by the function in a variable.
• Again all the divisorcount() function for a number less than the previous one store the value returned by the function in a variable.
• Compare both the variables.
• If they are equal print them.
• The divisorcount() function counts the number of divisors for a number.

Program:

// Header file
#include<stdio.h>

int divisorcount(int);

//Main function
int main()
{
    //Declare variables
    int n,divisor1,divisor2,j=0;

    //for-loop to call the function several times
    for(n=1;n<=107;n++){
        
        //Call the function for a number to count the number of divisors
        divisor1=divisorcount(n);
        
        //Call the same function for a number less than previous one
        divisor2=divisorcount(n-1);
        
        //Compare the numbers
        if(divisor1==divisor2){
            printf("%d and %d\n",n-1,n);
            j++;
        }
    }
    printf("\nThere are %d integers.",j);

}

//Function to count the number of divisors
int divisorcount(int num)
{
    int i,divi=0;
    for(i=1;i<=(num)/2;i++)
        if(num%i==0)
            divi++;
            
    return divi;
}