# Represent a number as a sum of maximum possible number of Prime Numbers

Given a positive integer . The task is to represent it as a sum of the maximum possible number of prime numbers. (N > 1)
Examples

```Input : N = 5
Output : 2 3

Input : N = 6
Output : 2 2 2```

At first, the problem might seem to involve some use of Goldbach’s conjecture. But the key observation here is to maximise the number of terms used, you should use as small numbers as possible. This leads to the following idea:

• If N is even, it can be represented as sum of two’s.
• Otherwise, has to be even and hence N can be represented as sum of one 3 and two’s.

This is the maximum number of primes whose sum is N.
Below is the implementation of the above approach:

## C++

 `// CPP program to represent a number as a` `// sum of maximum possible number of` `// Prime Numbers` `#include ` `using` `namespace` `std;`   `// Function to represent a number as a` `// sum of the maximum possible number` `// of Prime Numbers` `void` `printAsMaximalPrimeSum(``int` `n)` `{` `    ``// If n is odd, print one 3` `    ``if` `(n % 2 == 1) {` `        ``cout << ``"3 "``;` `        ``n -= 3;` `    ``}`   `    ``// Now n is even, print 2 n/2 times` `    ``while` `(n) {` `        ``cout << ``"2 "``;` `        ``n -= 2;` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `n = 5;` `    ``printAsMaximalPrimeSum(n);` `}`

## Java

 `// Java program to represent a number as a` `// sum of maximum possible number of` `// Prime Numbers`   `import` `java.io.*;`   `class` `GFG {` `   `  `// Function to represent a number as a` `// sum of the maximum possible number` `// of Prime Numbers` `static` `void` `printAsMaximalPrimeSum(``int` `n)` `{` `    ``// If n is odd, print one 3` `    ``if` `(n % ``2` `== ``1``) {` `        ``System.out.print( ``"3 "``);` `        ``n -= ``3``;` `    ``}`   `    ``// Now n is even, print 2 n/2 times` `    ``while` `(n>``0``) {` `        ``System.out.print( ``"2 "``);` `        ``n -= ``2``;` `    ``}` `}`   `       ``// Driver Code` `    ``public` `static` `void` `main (String[] args) {` `    ``int` `n = ``5``;` `    ``printAsMaximalPrimeSum(n);` `    ``}` `}`   `// This Code is contributed by inder_verma..`

## Python3

 `# Python3 program to represent a number as a` `# sum of maximum possible number of` `# Prime Numbers`     `# Function to represent a number as a` `# sum of the maximum possible number` `# of Prime Numbers` `def` `printAsMaximalPrimeSum( n):` ` `  `    ``# If n is odd, print one 3` `    ``if` `( n ``%` `2` `=``=` `1``):  ` `        ``print``(``"3 "``,end``=``"") ` `        ``n ``-``=` `3` `     `    `    ``# Now n is even, print 2 n/2 times` `    ``while` `( n>``0``):  ` `        ``print``(``"2 "``,end``=``"")` `        ``n ``-``=` `2` `     `  ` `  `    `  `# Driver Code`   `n ``=` `5` `printAsMaximalPrimeSum( n) `   `# This code is contributed by ihritik`

## C#

 `// C# program to represent a number as a` `// sum of maximum possible number of` `// Prime Numbers`     `using` `System;` `class` `GFG` `{` `    ``// Function to represent a number as a` `    ``// sum of the maximum possible number` `    ``// of Prime Numbers` `    ``static` `void` `printAsMaximalPrimeSum(``int` `n)` `    ``{` `        ``// If n is odd, print one 3` `        ``if` `(n % 2 == 1) {` `            ``Console.Write(``"3 "``);` `            ``n -= 3;` `        ``}` `    `  `        ``// Now n is even, print 2 n/2 times` `        ``while` `(n>0) {` `            ``Console.Write(``"2 "``);` `            ``n -= 2;` `        ``}` `    ``}` `    `  `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 5;` `        ``printAsMaximalPrimeSum(n);` `    ``}`   `}`   `// This code is contributed by ihritik`

## PHP

 `0) {` `        ``echo` `"2 "``;` `        ``\$n` `-= 2;` `    ``}` `}` `    `  `// Driver Code`   `\$n` `= 5;` `printAsMaximalPrimeSum(``\$n``);`     `// This code is contributed by ihritik` `?>`

## Javascript

 ``

Output:

`3 2`

Time Complexity: O(N)

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