# Break a number such that sum of maximum divisors of all parts is minimum

Last Updated : 23 Jun, 2022

We need to split a number n such that sum of maximum divisors of all the parts is minimum.

Examples :

```Input:  n = 27
Output: Minimum sum of maximum
divisors of parts = 3
Explanation : We can split 27 as
follows: 27 =  13 + 11 + 3,
Maximum divisor of 13 = 1,
11 = 1,
3 = 1.
Answer = 3 (1 + 1 + 1).

Input : n = 4
Output: Minimum sum of maximum
divisors of parts = 2
Explanation : We can write 4 = 2 + 2
Maximum divisor of 2 = 1,
So answer is 1 + 1 = 2.```

We need to minimize maximum divisors. It is obvious that if N is prime, maximum divisor = 1. If the number is not a prime, then the number should be atleast 2.

According to Goldbach’s Conjecture, every even integer can be expressed as sum of two prime numbers. For our problem there will be two cases:

1) When the number is even, it can be expressed as the sum of two prime numbers and our answer will be 2 because maximum divisor of a prime number is 1.
2) When the number is odd, it can also be written as sum of prime numbers, n = 2 + (n-2); if (n-2) is a prime number(answer = 2), otherwise. Refer odd number as sum of primes for details.
n = 3 + (n-3); (n-3) is an even number and it is sum of two primes(answer = 3).

Below is the implementation of this approach.

## C++

 `// CPP program to break a number such` `// that sum of maximum divisors of all` `// parts is minimum` `#include ` `using` `namespace` `std;`   `// Function to check if a number is` `// prime or not.` `bool` `isPrime(``int` `n)` `{` `    ``int` `i = 2;` `    ``while` `(i * i <= n) {` `        ``if` `(n % i == 0)` `            ``return` `false``;` `        ``i++;` `    ``}` `    ``return` `true``;` `}`   `int` `minimumSum(``int` `n)` `{` `    ``if` `(isPrime(n))` `        ``return` `1;`   `    ``// If n is an even number (we can` `    ``// write it as sum of two primes)` `    ``if` `(n % 2 == 0)` `        ``return` `2;`   `    ``// If n is odd and n-2 is prime.` `    ``if` `(isPrime(n - 2))` `        ``return` `2;`   `    ``// If n is odd, n-3 must be even.` `    ``return` `3;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 27;` `    ``cout << minimumSum(n);` `    ``return` `0;` `}`

## Java

 `// JAVA Code for Break a number such that sum` `// of maximum divisors of all parts is minimum` `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to check if a number is` `    ``// prime or not.` `    ``static` `boolean` `isPrime(``int` `n)` `    ``{` `        ``int` `i = ``2``;`   `        ``while` `(i * i <= n) {` `            ``if` `(n % i == ``0``)` `                ``return` `false``;` `            ``i++;` `        ``}` `        ``return` `true``;` `    ``}`   `    ``static` `int` `minimumSum(``int` `n)` `    ``{` `        ``if` `(isPrime(n))` `            ``return` `1``;`   `        ``// If n is an even number (we can` `        ``// write it as sum of two primes)` `        ``if` `(n % ``2` `== ``0``)` `            ``return` `2``;`   `        ``// If n is odd and n-2 is prime.` `        ``if` `(isPrime(n - ``2``))` `            ``return` `2``;`   `        ``// If n is odd, n-3 must be even.` `        ``return` `3``;` `    ``}`   `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``4``;` `        ``System.out.println(minimumSum(n));` `    ``}` `}`   `// This code is contributed by Arnav Kr. Mandal.`

## Python3

 `# Python3 program to break a number ` `# such that sum of maximum divisors ` `# of all parts is minimum`   `# Function to check if a number is ` `# prime or not.` `def` `isPrime( n):` `    ``i ``=` `2` `    `  `    ``while` `(i ``*` `i <``=` `n):` `        `  `        ``if` `(n ``%` `i ``=``=` `0``):` `            ``return` `False` `        ``i``+``=` `1`   `    ``return` `True` `    `  `def` `minimumSum( n):` `    `  `    ``if` `(isPrime(n)):` `        ``return` `1` `    `  `    ``# If n is an even number (we can` `    ``# write it as sum of two primes)` `    ``if` `(n ``%` `2` `=``=` `0``):` `        ``return` `2` `    `  `    ``# If n is odd and n-2 is prime.` `    ``if` `(isPrime(n ``-` `2``)):` `        ``return` `2` `        `  `    ``# If n is odd, n-3 must be even.` `    ``return` `3`   `# Driver code` `n ``=` `27` `print``(minimumSum(n))`   `# This code is contributed by "Abhishek Sharma 44"`

## C#

 `// C# Code for Break a number` `// such that sum of maximum` `// divisors of all parts is minimum` `using` `System;`   `class` `GFG {`   `    ``// Function to check if a number is` `    ``// prime or not.` `    ``static` `bool` `isPrime(``int` `n)` `    ``{` `        ``int` `i = 2;`   `        ``while` `(i * i <= n) {` `            ``if` `(n % i == 0)` `                ``return` `false``;` `            ``i++;` `        ``}` `        ``return` `true``;` `    ``}`   `    ``static` `int` `minimumSum(``int` `n)` `    ``{` `        ``if` `(isPrime(n))` `            ``return` `1;`   `        ``// If n is an even number (we can` `        ``// write it as sum of two primes)` `        ``if` `(n % 2 == 0)` `            ``return` `2;`   `        ``// If n is odd and n-2 is prime.` `        ``if` `(isPrime(n - 2))` `            ``return` `2;`   `        ``// If n is odd, n-3 must be even.` `        ``return` `3;` `    ``}`   `    ``// Driver program` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 27;` `        ``Console.WriteLine(minimumSum(n));` `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

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## Javascript

Output :

`    3`

Time Complexity: O(?n)
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.