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Find third number such that sum of all three number becomes prime
  • Last Updated : 18 Mar, 2021

Given two numbers A and B. The task is to find the smallest positive integer greater than or equal to 1 such that the sum of all three numbers become a prime number.
Examples: 
 

Input: A = 2, B = 3 
Output:
Explaination: 
The third number is 2 because if we add all three number the 
sum obtained is 7 which is a prime number.
Input: A = 5, B = 3 
Output:
 

 

Approach: 
 

  1. First of all store the sum of given 2 number in sum variable.
  2. Now, check if bitwise and (&) of sum and 1 is equal to 1 or not (checking if the number is odd or not).
  3. If it is equal to 1 then assign 2 to variable temp and go to step 4.
  4. Else check sum value of sum and temp variable whether it is prime number or not. If prime, then print the value of temp variable else add 2 to temp variable untill it is less than a prime value.

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that will check
// whether number is prime or not
bool prime(int n)
{
    for (int i = 2; i * i <= n; i++) {
 
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to print the 3rd number
void thirdNumber(int a, int b)
{
    int sum = 0, temp = 0;
 
    sum = a + b;
    temp = 1;
 
    // If the sum is odd
    if (sum & 1) {
        temp = 2;
    }
 
    // If sum is not prime
    while (!prime(sum + temp)) {
        temp += 2;
    }
 
    cout << temp;
}
 
// Driver code
int main()
{
    int a = 3, b = 5;
 
    thirdNumber(a, b);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
     
class GFG
{
     
// Function that will check
// whether number is prime or not
static boolean prime(int n)
{
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to print the 3rd number
static void thirdNumber(int a, int b)
{
    int sum = 0, temp = 0;
 
    sum = a + b;
    temp = 1;
 
    // If the sum is odd
    if (sum == 0)
    {
        temp = 2;
    }
 
    // If sum is not prime
    while (!prime(sum + temp))
    {
        temp += 2;
    }
    System.out.print(temp);
}
 
// Driver code
static public void main (String []arr)
{
    int a = 3, b = 5;
     
    thirdNumber(a, b);
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the above approach
 
# Function that will check
# whether number is prime or not
def prime(n):
    for i in range(2, n + 1):
        if i * i > n + 1:
            break
 
        if (n % i == 0):
            return False
 
    return True
 
# Function to print the 3rd number
def thirdNumber(a, b):
    summ = 0
    temp = 0
 
    summ = a + b
    temp = 1
 
    #If the summ is odd
    if (summ & 1):
        temp = 2
 
    #If summ is not prime
    while (prime(summ + temp) == False):
        temp += 2
 
    print(temp)
 
# Driver code
a = 3
b = 5
 
thirdNumber(a, b)
 
# This code is contributed by Mohit Kumar

C#




// C# implementation of the above approach
using System;
 
class GFG
{
     
// Function that will check
// whether number is prime or not
static bool prime(int n)
{
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to print the 3rd number
static void thirdNumber(int a, int b)
{
    int sum = 0, temp = 0;
 
    sum = a + b;
    temp = 1;
 
    // If the sum is odd
    if (sum == 0)
    {
        temp = 2;
    }
 
    // If sum is not prime
    while (!prime(sum + temp))
    {
        temp += 2;
    }
    Console.Write (temp);
}
 
// Driver code
static public void Main ()
{
    int a = 3, b = 5;
     
    thirdNumber(a, b);
}
}
 
// This code is contributed by Sachin.

Javascript




<script>
    // Javascript implementation of the above approach
     
    // Function that will check
    // whether number is prime or not
    function prime(n)
    {
        for (let i = 2; i * i <= n; i++) {
 
            if (n % i == 0)
                return false;
        }
        return true;
    }
 
    // Function to print the 3rd number
    function thirdNumber(a, b)
    {
        let sum = 0, temp = 0;
 
        sum = a + b;
        temp = 1;
 
        // If the sum is odd
        if ((sum & 1) != 0) {
            temp = 2;
        }
 
        // If sum is not prime
        while (!prime(sum + temp)) {
            temp += 2;
        }
        document.write(temp);
    }
     
    let a = 3, b = 5;
    thirdNumber(a, b);
 
// This code is contributed by divyeshrabadiya07.
</script>
Output: 
3

 

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