# Checking for Prime number using recursion

## Checking for Prime number using recursion

Given an input number , write a program for Checking for Prime number using recursion .

**input:-**

23

**Output:-**

23 is a prime number.

## Checking for Prime number using recursion:-

The objective of this code is to recursively check that a given number is prime or not . So for this we will write a function CheckPrime() that will take two parameter number and **i ( i for checking the factors of the number ). ** we initialize the i with 2 . The base case for the function is if number is equal to i then it is a prime number so we will return 1 . otherwise we check that if number is divisible by i then it is not a prime number so return 0 and if not divisible then increase the value of i by 1 and call the CheckPrime() function recursively . For better understanding see the example :-

**Code (C ++):-**

#include<bits/stdc++.h> using namespace std; // recursive function to check number is // prime or not int CheckPrime(int n ,int i) { if(n==i) return 1; else { if(n%i==0) return 0; else return CheckPrime(n,i+1); } } int main() { int n; cout<<"Enter a number"<<endl; cin>>n; if(CheckPrime(n,2)==1) cout<<n<<" is a prime number"<<endl; else cout<<n<<" is not a prime number"<<endl; return 0; }

**Output:-**

Enter a number 23 23 is a prime number

python code:-

def prime(x,i): if x==i: return 1 elif x%i==0: return 0 else: return prime(x,i+1) n=int(input("Enter a number ")) n1=prime(n,2) if n1==1: print(n,"is a prime number") else: print(n,"is not a prime number")

output:-

Enter a number 11 11 is a prime number

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