Pointer-Prime Number is a prime number p such that the next prime after p can be obtained from p by adding the product of the digits of p.
Some Pointer primes are:
23, 61, 1123, 1231, 1321, 2111, 2131, 11261….
Check if N is a Pointer prime number
Given a number N, the task is to check if N is a Pointer-Prime Number or not. If N is a Pointer-Prime Number then print “Yes” else print “No”.
Examples:
Input: N = 23
Output: Yes
Explanation:
23 + product of digits of 23 = 29,
which is the next prime after 23.
Input: N = 29
Output: No
Approach: :
- Find the product of digits of N
- Then, find the next prime number to N
- Now if N is prime and N + product of digits of N equals next prime to N, then print “Yes” else print “No”.
Below is the implementation of the above approach:
C++
// C++ implementation for the// above approach#include <bits/stdc++.h>using namespace std;// Function to find the product of// digits of a number Nint digProduct(int n){ int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product;}// Function that returns true if n// is prime else returns falsebool isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to return the smallest// prime number greater than Nint nextPrime(int N){ // Base case if (N <= 1) return 2; int prime = N; bool found = false; // Loop continuously until isPrime returns // true for a number greater than n while (!found) { prime++; if (isPrime(prime)) found = true; } return prime;}// Function to check Pointer-Prime numbersbool isPointerPrime(int n){ if (isPrime(n) && (n + digProduct(n) == nextPrime(n))) return true; else return false;}// Driver Codeint main(){ // Given Number N int N = 23; // Function Call if (isPointerPrime(N)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program for above approachclass GFG{ // Function to find the product of// digits of a number Nstatic int digProduct(int n){ int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product;}// Function that returns true if n// is prime else returns falsestatic boolean isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to return the smallest// prime number greater than Nstatic int nextPrime(int N){ // Base case if (N <= 1) return 2; int prime = N; boolean found = false; // Loop continuously until isPrime returns // true for a number greater than n while (!found) { prime++; if (isPrime(prime)) found = true; } return prime;}// Function to check Pointer-Prime numbersstatic boolean isPointerPrime(int n){ if (isPrime(n) && (n + digProduct(n) == nextPrime(n))) return true; else return false;}// Driver Code public static void main(String[] args) { // Given Number N int N = 23; // Function Call if (isPointerPrime(N)) System.out.print("Yes"); else System.out.print("No");} } // This code is contributed by Shubham Prakash |
Python3
# Python3 implementation for the above approach def digProduct(n): product = 1 while(n != 0): product = product * (n % 10) n = int(n / 10) return product# Function that returns true if n# is prime else returns false def isPrime(n): # Corner cases if (n <= 1): return False if (n <= 3): return True # This is checked so that we can skip # middle five numbers in below loop if (n % 2 == 0 or n % 3 == 0): return False i = 5 while(i * i <= n): if (n % i == 0 or n % (i + 2) == 0): return False i = i + 6 return True# Function to return the smallest prime# number greater than N def nextPrime(N): # Base case if(N <= 1): return 2; prime = N found = False # Loop continuously until isPrime # returns true for a number greater # than n while(not found): prime = prime + 1 if(isPrime(prime)): found = True return prime# Function to check Pointer-Prime numbers def isPointerPrime(n): if(isPrime(n) and (n + digProduct(n) == nextPrime(n))): return True else: return False# Driver Codeif __name__=="__main__": # Given number N N = 23 # Function call if(isPointerPrime(N)): print("Yes") else: print("No")# This code is contributed by adityakumar27200 |
C#
// C# program for above approachusing System;class GFG{ // Function to find the product of// digits of a number Nstatic int digProduct(int n){ int product = 1; while (n != 0) { product = product * (n % 10); n = n / 10; } return product;}// Function that returns true if n// is prime else returns falsestatic bool isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to return the smallest// prime number greater than Nstatic int nextPrime(int N){ // Base case if (N <= 1) return 2; int prime = N; bool found = false; // Loop continuously until isPrime returns // true for a number greater than n while (!found) { prime++; if (isPrime(prime)) found = true; } return prime;}// Function to check Pointer-Prime numbersstatic bool isPointerPrime(int n){ if (isPrime(n) && (n + digProduct(n) == nextPrime(n))) return true; else return false;}// Driver Code public static void Main() { // Given Number N int N = 23; // Function Call if (isPointerPrime(N)) Console.Write("Yes"); else Console.Write("No");} } // This code is contributed by Code_Mech |
Javascript
<script>// Javascript implementation for the// above approach// Function to find the product of// digits of a number Nfunction digProduct(n){ var product = 1; while (n != 0) { product = product * (n % 10); n = parseInt(n / 10); } return product;}// Function that returns true if n// is prime else returns falsefunction isPrime(n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (var i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to return the smallest// prime number greater than Nfunction nextPrime(N){ // Base case if (N <= 1) return 2; var prime = N; var found = false; // Loop continuously until isPrime returns // true for a number greater than n while (!found) { prime++; if (isPrime(prime)) found = true; } return prime;}// Function to check Pointer-Prime numbersfunction isPointerPrime(n){ if (isPrime(n) && (n + digProduct(n) == nextPrime(n))) return true; else return false;}// Driver Code// Given Number Nvar N = 23;// Function Callif (isPointerPrime(N)) document.write( "Yes");else document.write( "No");</script> |
Output:
Yes
Time Complexity: O(n).
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