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Maximum Primes whose sum is equal to given N

  • Last Updated : 29 Apr, 2021

Given a positive integer N > 1. Find the maximum count of prime numbers whose sum is equal to given N.

Examples: 

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Input : N = 5 
Output :
Explanation : 2 and 3 are two prime numbers whose sum is 5.
Input : N = 6 
Output :
Explanation : 2, 2, 2 are three prime numbers whose sum is 6.

For the maximum number of primes whose sum is equal to given n, prime numbers must be as small as possible. So, 2 is the smallest possible prime number and is an even number. The next prime number greater than 2 is 3 which is odd. So, for any given n there are two conditions, either n will be odd or even. 

  • Case 1: n is even, In this case, n/2 will be the answer (n/2 number of 2 will result in the sum of n). 
     
  • Case 2: n is odd, In this case, floor(n/2) will be the answer ((n-3)/2 number of 2, and one 3 will result in the sum of n). 
     

Below is the implementation of the above approach: 
 

C++




// C++ program for above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find max count of primes
int maxPrimes(int n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2)  = (int)(n/2) is required answer
    return n / 2;
}
 
// Driver Code
int main()
{
    int n = 17;
 
    cout << maxPrimes(n);
 
    return 0;
}

Java




// Java program for above approach
class GFG
{
 
// Function to find max count of primes
static int maxPrimes(int n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2) = (int)(n/2)
    // is required answer
    return n / 2;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 17;
 
    System.out.println(maxPrimes(n));
}
}
 
// This code is contributed
// by Code_Mech

Python3




# Python3 program for above approach
 
# Function to find max count of primes
def maxPrmimes(n):
 
    # if n is even n/2 is required answer
    # if n is odd floor(n/2) = (int)(n/2)
    # is required answer
    return n // 2
 
# Driver code
n = 17
print(maxPrmimes(n))
 
# This code is contributed
# by Shrikant13

C#




// C# program for above approach
using System;
 
class GFG
{
 
// Function to find max count of primes
static int maxPrimes(int n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2) = (int)(n/2)
    // is required answer
    return n / 2;
}
 
// Driver Code
public static void Main()
{
    int n = 17;
 
    Console.WriteLine(maxPrimes(n));
}
}
 
// This code is contributed
// by Akanksha Rai

PHP




<?php
// PHP program for above approach
 
// Function to find max count of primes
function maxPrimes($n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2) = (int)(n/2) is required answer
    return (int)($n / 2);
}
 
    // Driver Code
    $n = 17;
    echo maxPrimes($n);
 
// This code is contributed by mits
?>

Javascript




<script>
 
 
// Javascript program for above approach
 
// Function to find max count of primes
function maxPrimes(n)
{
    // if n is even n/2 is required answer
    // if n is odd floor(n/2)  = (int)(n/2) is required answer
    return parseInt(n / 2);
}
 
// Driver Code
var n = 17;
document.write( maxPrimes(n)); 
 
</script>   
Output: 
8

 




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