Right-Truncatable Prime

A Right-truncatable prime is a prime which remains prime when the last (“right”) digit is successively removed. For example, 239 is right-truncatable prime since 239, 23 and 2 are all prime. There are 83 right-truncatable primes.

The task is to check whether the given number (N > 0) is right-truncatable prime or not.

Examples:

Input: 239
Output: Yes

Input: 101
Output: No
101 is not right-truncatable prime because 
numbers formed are 101, 10 and 1. Here, 101 
is prime but 10 and 1 are not prime.



The idea is to generate all the primes less than or equal to the given number N using Sieve of Eratosthenes. Once we have generated all such primes, then we check whether the number remains prime when the last (“right”) digit is successively removed.

CPP

// Program to check 
// whether a given number
// is right-truncatable 
// prime or not.
#include<bits/stdc++.h>
using namespace std;

// Generate all prime numbers less than n.
bool sieveOfEratosthenes(int n, bool isPrime[])
{
    // Initialize all entries
    // of boolean array as 
    // true. A value in
    // isPrime[i] will finally
    // be false if i is Not a 
    // prime, else true
    // bool isPrime[n+1];
    isPrime[0] = isPrime[1] = false;
    for( int i = 2; i <= n; i++)
        isPrime[i] = true;

    for (int p = 2; p * p<=n; p++)
    {

        // If isPrime[p] is not changed, then it is
        // a prime
        if (isPrime[p] == true)
        {
            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                isPrime[i] = false;

        }
    }
}

// Returns true if n is right-truncatable,
// else false
bool rightTruPrime(int n)
{
    // Generating primes using Sieve
    bool isPrime[n+1];
    sieveOfEratosthenes(n, isPrime);

    // Checking whether the number remains
    // prime when the last ("right") 
    // digit is successively removed
    while (n)
    {
        if (isPrime[n]) 
            n = n / 10;
        else
            return false;
    }
    return true; 
}

// Driver program
int main()
{
    int n = 59399;
    if (rightTruPrime(n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java

// Java code to check 
// right-truncatable 
// prime or not.
import java.io.*;

class GFG {
    
    // Generate all prime
    // numbers less than n.
    static void sieveOfEratosthenes
                (int n, boolean isPrime[]) 
    {
        
        // Initialize all entries of
        // boolean array as true. A
        // value in isPrime[i] will
        // finally be false if i is
        // Not a prime, else true
        // bool isPrime[n+1];
        isPrime[0] = isPrime[1] = false;
        for (int i = 2; i <= n; i++)
            isPrime[i] = true;
    
        for (int p=2; p*p<=n; p++)
        {
            // If isPrime[p] is not 
            // changed, then it 
            // is a prime
            if (isPrime[p] == true)
            {
                // Update all multiples of p
                for (int i = p * 2; i <= n; i += p)
                    isPrime[i] = false;
            }
        }
    }
    
    // Returns true if n is
    // right-truncatable, 
    // else false
    static boolean rightTruPrime(int n)
     {
        
        // Generating primes using Sieve
        boolean isPrime[] = new boolean[n+1];
        sieveOfEratosthenes(n, isPrime);
    
        // Checking whether the number
        // remains prime when the last (right)
        // digit is successively removed
        while (n != 0)
        {
            
            if (isPrime[n])
                n = n / 10; 
            else
                return false;
        }
        return true;
    }
    
    // Driver program
    public static void main(String args[])
    {
        int n = 59399;
        
        if (rightTruPrime(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

/* This code is contributed by Nikita Tiwari.*/

Python3

# Python3 Program to check
# whether a given number
# is right-truncatable 
# prime or not.

# Generate all prime numbers less than n.
def sieveOfEratosthenes(n,isPrime) :
    
    # Initialize all entries 
    # of boolean array as 
    # true. A value in isPrime[i] 
    # will finally be false if
    # i is Not a prime, else true
    # bool isPrime[n+1];
    isPrime[0] = isPrime[1] = False
    for i in range(2, n+1) :
        isPrime[i] = True
    p = 2
    while(p * p <= n) :
        # If isPrime[p] is not changed, then it is
        # a prime
        if (isPrime[p] == True) :
            # Update all multiples of p
            i = p * 2
            while(i <= n) :
                isPrime[i] = False
                i = i + p
        p = p + 1
        

# Returns true if n is right-truncatable, else false
def rightTruPrime(n) :
    # Generating primes using Sieve
    isPrime=[None] * (n+1)
    sieveOfEratosthenes(n, isPrime)

    # Checking whether the 
    # number remains prime
    # when the last ("right")
    # digit is successively
    # removed
    while (n != 0) :
        if (isPrime[n]) :
            n = n // 10     
        else :
            return False 
    
    return True


# Driven program
n = 59399
if (rightTruPrime(n)) :
    print("Yes")
else :
    print("No")

# This code is contributed by Nikita Tiwari.

C#


// C# code to check right-
// truncatable prime or not
using System;

class GFG {

    // Generate all prime
    // numbers less than n.
    static void sieveOfEratosthenes(int n, bool[] isPrime)
    {

        // Initialize all entries of
        // boolean array as true. A
        // value in isPrime[i] will
        // finally be false if i is
        // Not a prime, else true
        // bool isPrime[n+1];
        isPrime[0] = isPrime[1] = false;

        for (int i = 2; i <= n; i++)
            isPrime[i] = true;

        for (int p = 2; p * p <= n; p++) {
            // If isPrime[p] is not
            // changed, then it
            // is a prime
            if (isPrime[p] == true) {
                // Update all multiples of p
                for (int i = p * 2; i <= n; i += p)
                    isPrime[i] = false;
            }
        }
    }

    // Returns true if n is right-
    // truncatable,  else false
    static bool rightTruPrime(int n)
    {

        // Generating primes using Sieve
        bool[] isPrime = new bool[n + 1];
        sieveOfEratosthenes(n, isPrime);

        // Checking whether the number
        // remains prime when last (right)
        // digit is successively removed
        while (n != 0) {

            if (isPrime[n])
                n = n / 10;
            else
                return false;
        }
        return true;
    }

    // Driven program
    public static void Main()
    {
        int n = 59399;

        if (rightTruPrime(n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

// This code is contributed by Anant Agarwal


Output:

Yes

Related Article:Left-Truncatable Prime

References:
https://en.wikipedia.org/wiki/Truncatable_prime

This article is contributed by Rahul Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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