# C++ program to find 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.

Example:

Input

• Number=30
• X=3

Output

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

## Algorithm

• 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 :";    cin>>Number;    cout<<"\nEnter exact number of divisors :";    cin>>X;    count1 = 0;    for(int i=1;i<=Number;i++)    {        int count2 = 0;        for(int j=1;j<=i;j++)        {            if(i%j==0)            {                count2++;            }        }        if(count2==X)        {            count1++;            cout<<i;        }    }    cout<<count1;    return 0;}`
`Enter range of number :30Enter exact number of divisors :349253`