Additive Prime Number
Last Updated :
26 Apr, 2023
Given a number N, the task is to check if N is an Additive Prime Number or not. If N is an Additive Prime Number then print “Yes” else print “No”.
An Additive Prime Number is a prime number P such that the sum of digits of P is also a prime number.
For example, 23 is an Additive prime because 2 + 3 = 5 which is a prime number.
Examples:
Input: N = 23
Output: Yes
Explanation: Sum of digits of 23 = 2 + 3 = 5.
Input: N = 10
Output: No
Approach: The idea is to find the sum of digits of the number N and check if it is prime or not. If the sum is a prime number then print “Yes” else print “No”.
Algorithm:
- Define a function named isPrime that takes an integer as input and returns true if it is a prime number, and false otherwise.
- Check if the input number is less than or equal to 1. If yes, return false.
- Use a for loop to iterate from 2 to the square root of the input number. If the input number is divisible by any number in this range, return false. Otherwise, return true.
- Define a function named digit_sum that takes an integer as input and returns the sum of its digits.
- Initialize a variable ans to 0.
- Use a while loop to iterate as long as the input number is greater than 0.
- Add the last digit of the input number to ans and remove the last digit from the input number.
- Return ans.
- Define a function named additive_prime that takes an integer as input and returns 1 if it is an additive prime, and 0 otherwise.
- Check if the input number is a prime number using the isPrime function.
- If the input number is a prime number, calculate the sum of its digits using the digit_sum function.
- Check if the sum of digits is a prime number using the isPrime function.
- If both the input number and its sum of digits are prime numbers, return 1. Otherwise, return 0.
- In the main function:
a. Define an integer variable.
b. Call the additive_prime function with the integer variable as an argument.
c. If the function returns 1, print “Yes” to the console. Otherwise, print “No”.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
bool isPrime(int n)
{
if (n <= 1)
return false;
for (int i = 2; i * i <= n; i++) {
if (n % i == 0)
return false;
}
return true;
}
int digit_sum(int n)
{
int ans = 0;
while (n > 0) {
ans += (n % 10);
n /= 10;
}
return ans;
}
int additive_prime(int n)
{
if (isPrime(n)) {
int d_sum = digit_sum(n);
if (isPrime(d_sum))
return 1;
}
return 0;
}
int main()
{
int n = 23;
if (additive_prime(n))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
|
Java
import java.util.*;
public class Gfg {
public static boolean isPrime(int n)
{
if (n <= 1) {
return false;
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public static int digit_sum(int n) {
int ans = 0;
while (n > 0) {
ans += (n % 10);
n /= 10;
}
return ans;
}
public static int additive_prime(int n) {
if (isPrime(n)) {
int d_sum = digit_sum(n);
if (isPrime(d_sum)) {
return 1;
}
}
return 0;
}
public static void main(String[] args) {
int n = 23;
if (additive_prime(n) == 1) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}
|
Python
def isPrime(n):
if (n <= 1):
return False
for i in range(2, int(n**0.5) + 1):
if (n % i == 0):
return False
return True
def digit_sum(n):
ans = 0
while (n > 0):
ans += (n % 10)
n = int(n/10)
return ans
def additive_prime(n):
if (isPrime(n)):
d_sum = digit_sum(n)
if (isPrime(d_sum)):
return 1
return 0
if __name__ == '__main__':
n = 23
if (additive_prime(n)):
print("Yes")
else:
print("No")
|
C#
using System;
namespace AdditivePrime
{
class Program
{
static bool IsPrime(int n)
{
if (n <= 1) {
return false;
}
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
static int DigitSum(int n)
{
int ans = 0;
while (n > 0) {
ans += (n % 10);
n /= 10;
}
return ans;
}
static bool IsAdditivePrime(int n)
{
if (IsPrime(n)) {
int dSum = DigitSum(n);
if (IsPrime(dSum)) {
return true;
}
}
return false;
}
static void Main(string[] args)
{
int n = 23;
if (IsAdditivePrime(n)) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
}
}
|
Javascript
function isPrime(n) {
if (n <= 1) {
return false;
}
for (let i = 2; i * i <= n; i++) {
if (n % i === 0) {
return false;
}
}
return true;
}
function digit_sum(n) {
let ans = 0;
while (n > 0) {
ans += n % 10;
n = Math.floor(n / 10);
}
return ans;
}
function additive_prime(n) {
if (isPrime(n)) {
const d_sum = digit_sum(n);
if (isPrime(d_sum)) {
return 1;
}
}
return 0;
}
const n = 23;
if (additive_prime(n)) {
console.log("Yes");
} else {
console.log("No");
}
|
Time complexity: O(logn)
Auxiliary Space: O(1), as constant space is used.
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