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

November 2, 2021

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 :30

Enter exact number of divisors :3
49253