Print all multiplicative primes <= N

Given an integer N, the task is to print all the multiplicative primes ≤ N.

Multiplicative Primes are the primes such that the product of their digits is also a prime. For example; 2, 3, 7, 13, 17, …

Examples:

Input: N = 10
Output: 2 3 5 7

Input: N = 3
Output: 2 3

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Using Sieve of Eratosthenes check for all the primes ≤ N whether they are multiplicative primes i.e. product of their digits is also a prime. If yes then print those multiplicative primes.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;    // Function to return the digit product of n int digitProduct(int n) {     int prod = 1;     while (n) {         prod = prod * (n % 10);         n = n / 10;     }        return prod; }    // Function to print all multiplicative primes <= n void printMultiplicativePrimes(int n) {     // Create a boolean array "prime[0..n+1]". A     // value in prime[i] will finally be false     // if i is Not a prime, else true.     bool prime[n + 1];     memset(prime, true, sizeof(prime));        prime = prime = false;     for (int p = 2; p * p <= n; p++) {            // If prime[p] is not changed, then         // it is a prime         if (prime[p]) {                // Update all multiples of p             for (int i = p * 2; i <= n; i += p)                 prime[i] = false;         }     }        for (int i = 2; i <= n; i++) {            // If i is prime and its digit sum is also prime         // i.e. i is a multiplicative prime         if (prime[i] && prime[digitProduct(i)])             cout << i << " ";     } }    // Driver code int main() {     int n = 10;     printMultiplicativePrimes(n); }

Java

 // Java implementation of the approach import java.io.*;    class GFG  {    // Function to return the digit product of n static int digitProduct(int n) {     int prod = 1;     while (n > 0)      {         prod = prod * (n % 10);         n = n / 10;     }     return prod; }    // Function to print all multiplicative primes <= n static void printMultiplicativePrimes(int n) {     // Create a boolean array "prime[0..n+1]". A     // value in prime[i] will finally be false     // if i is Not a prime, else true.     boolean prime[] = new boolean[n + 1 ];     for(int i = 0; i <= n; i++)      prime[i] = true;        prime = prime = false;     for (int p = 2; p * p <= n; p++)      {            // If prime[p] is not changed, then         // it is a prime         if (prime[p])          {                // Update all multiples of p             for (int i = p * 2; i <= n; i += p)                 prime[i] = false;         }     }        for (int i = 2; i <= n; i++)     {            // If i is prime and its digit sum is also prime         // i.e. i is a multiplicative prime         if (prime[i] && prime[digitProduct(i)])             System.out.print( i + " ");     } }        // Driver code     public static void main (String[] args)      {         int n = 10;         printMultiplicativePrimes(n);     } }    // This code is contributed by shs..

Python3

 # Python 3 implementation of the approach from math import sqrt    # Function to return the digit product of n def digitProduct(n):     prod = 1     while (n):         prod = prod * (n % 10)         n = int(n / 10)        return prod    # Function to print all multiplicative # primes <= n def printMultiplicativePrimes(n):            # Create a boolean array "prime[0..n+1]".      # A value in prime[i] will finally be      # false if i is Not a prime, else true.     prime = [True for i in range(n + 1)]        prime = prime = False     for p in range(2, int(sqrt(n)) + 1, 1):                    # If prime[p] is not changed,          # then it is a prime         if (prime[p]):                            # Update all multiples of p             for i in range(p * 2, n + 1, p):                 prime[i] = False                for i in range(2, n + 1, 1):                    # If i is prime and its digit sum          # is also prime i.e. i is a         # multiplicative prime         if (prime[i] and prime[digitProduct(i)]):             print(i, end = " ")    # Driver code if __name__ == '__main__':     n = 10     printMultiplicativePrimes(n)    # This code is contributed by # Surendra_Gangwar

C#

 // C# implementation of the approach class GFG  {    // Function to return the digit product of n static int digitProduct(int n) {     int prod = 1;     while (n > 0)      {         prod = prod * (n % 10);         n = n / 10;     }     return prod; }    // Function to print all multiplicative primes <= n static void printMultiplicativePrimes(int n) {     // Create a boolean array "prime[0..n+1]". A     // value in prime[i] will finally be false     // if i is Not a prime, else true.     bool[] prime = new bool[n + 1 ];            for(int i = 0; i <= n; i++)         prime[i] = true;        prime = prime = false;     for (int p = 2; p * p <= n; p++)      {            // If prime[p] is not changed, then         // it is a prime         if (prime[p])          {                // Update all multiples of p             for (int i = p * 2; i <= n; i += p)                 prime[i] = false;         }     }        for (int i = 2; i <= n; i++)     {            // If i is prime and its digit sum is also prime         // i.e. i is a multiplicative prime         if (prime[i] && prime[digitProduct(i)])             System.Console.Write( i + " ");     } }        // Driver code     static void Main()      {         int n = 10;         printMultiplicativePrimes(n);     } }    // This code is contributed by chandan_jnu

PHP



Output:

2 3 5 7

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