C++ program to find number of integers which has exactly x divisors

Number of Integers which has exactly X Divisors

Number of integers which has exactly X divisors

In this page we will learn how to find number of integers which has exactly x divisors. Divisors are numbers which  perfectly divides a number. For example take 6 as a number then divisors of 6 are 1,2,3,6. ‘1’ and number itself are always divisors of the same number. If a number has exactly 2 divisors then that number is a prime number.

Here our task is to write a  C program to count out how many number of integers which has exactly x divisors. This can be done by finding number of divisors that each number has and then checking with the given ‘X‘. If X and number of divisors match then we will increment the counter value.



  • Number=30
  • X=3


  • numbers with exactly 3 divisors are 4 9 25
  • count= 3


  • Take user inputs like Number and X.
  • Initialize a count variable with zero value.
  • Run a for loop with a range from 1 to Number+1.
  • Initialize another count variable with zero.
  •  Run other for loop ranging from 1 to iterator of 1st for loop+1.
  •  Check for complete division conditions and if TRUE increment count2 by 1.
  •  Come out of for loop and check if count2 is equal to Divisor.
  •  If TRUE increment count1 by 1 and print the number with exact divisors.
  • Print count1.

C++ Code:-

#include <iostream>

using namespace std;

int main()
int Number,X,count1;
cout<<"Enter range of number :";
cout<<"\nEnter exact number of divisors :";
count1 = 0;
for(int i=1;i<=Number;i++)
int count2 = 0;
for(int j=1;j<=i;j++)
return 0;
Enter range of number :30

Enter exact number of divisors :3