Related Articles

# Sum of all second largest divisors after splitting a number into one or more parts

• Difficulty Level : Medium
• Last Updated : 01 Apr, 2021

Given an integer N( 2 <= N <= 10^9 ), split the number into one or more parts(possibly none), where each part must be greater than 1. The task is to find the minimum possible sum of the second largest divisor of all the splitting numbers.
Examples:

```Input : N = 27
Output : 3
Explanation : Split the given number into 19, 5, 3. Second largest
divisor of each number is 1. So, sum is 3.

Input : N = 19
Output : 1
Explanation : Don't make any splits. Second largest divisor of 19
is 1. So, sum is 1 ```

Approach:
The idea is based on Goldbach’s conjecture

1. When the number is prime, then the answer will be 1.
2. When a number is even then it can always be expressed as a sum of 2 primes. So, the answer will be 2.
3. When the number is odd,
• When N-2 is prime, then the number can be express as the sum of 2 primes, that are 2 and N-2, then the answer will be 2.
• Otherwise, the answer will always be 3.

Below is the implementation of the above approach:

## C++

 `// CPP program to find sum of all second largest divisor``// after splitting a number into one or more parts``#include ``using` `namespace` `std;` `// Function to find a number is prime or not``bool` `prime(``int` `n)``{``    ``if` `(n == 1)``        ``return` `false``;` `    ``// If there is any divisor``    ``for` `(``int` `i = 2; i * i <= n; ++i)``        ``if` `(n % i == 0)``            ``return` `false``;` `    ``return` `true``;``}` `// Function to find the sum of all second largest divisor``// after splitting a number into one or more parts``int` `Min_Sum(``int` `n)``{``    ``// If number is prime``    ``if` `(prime(n))``        ``return` `1;` `    ``// If n is even``    ``if` `(n % 2 == 0)``        ``return` `2;` `    ``// If the number is odd``    ``else` `{` `        ``// If N-2 is prime``        ``if` `(prime(n - 2))``            ``return` `2;` `        ``// There exists 3 primes x1, x2, x3``        ``// such that x1 + x2 + x3 = n``        ``else``            ``return` `3;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `n = 27;` `    ``// Function call``    ``cout << Min_Sum(n);` `    ``return` `0;``}`

## Java

 `// Java program to Sum of all second largest``// divisors after splitting a number into one or more parts``import` `java.io.*;`` ` `class` `GFG {`` ` ` ` ` ` `// Function to find a number is prime or not``static` `boolean` `prime(``int` `n)``{``    ``if` `(n == ``1``)``        ``return` `false``;`` ` `    ``// If there is any divisor``    ``for` `(``int` `i = ``2``; i * i <= n; ++i)``        ``if` `(n % i == ``0``)``            ``return` `false``;`` ` `    ``return` `true``;``}`` ` `// Function to find the sum of all second largest divisor``// after splitting a number into one or more parts``static` `int` `Min_Sum(``int` `n)``{``    ``// If number is prime``    ``if` `(prime(n))``        ``return` `1``;`` ` `    ``// If n is even``    ``if` `(n % ``2` `== ``0``)``        ``return` `2``;`` ` `    ``// If the number is odd``    ``else` `{`` ` `        ``// If N-2 is prime``        ``if` `(prime(n - ``2``))``            ``return` `2``;`` ` `        ``// There exists 3 primes x1, x2, x3``        ``// such that x1 + x2 + x3 = n``        ``else``            ``return` `3``;``    ``}``}`` ` `// Driver code`` ` ` ` `    ``public` `static` `void` `main (String[] args) {``    ``int` `n = ``27``;`` ` `    ``// Function call``    ``System.out.println( Min_Sum(n));``    ``}``}` `// This code is contributed by anuj_6`

## Python3

 `# Python 3 program to find sum of all second largest divisor``# after splitting a number into one or more parts` `from` `math ``import` `sqrt``# Function to find a number is prime or not``def` `prime(n):``    ``if` `(n ``=``=` `1``):``        ``return` `False` `    ``# If there is any divisor``    ``for` `i ``in` `range``(``2``,``int``(sqrt(n))``+``1``,``1``):``        ``if` `(n ``%` `i ``=``=` `0``):``            ``return` `False` `    ``return` `True` `# Function to find the sum of all second largest divisor``# after splitting a number into one or more parts``def` `Min_Sum(n):``    ``# If number is prime``    ``if` `(prime(n)):``        ``return` `1` `    ``# If n is even``    ``if` `(n ``%` `2` `=``=` `0``):``        ``return` `2` `    ``# If the number is odd``    ``else``:``        ``# If N-2 is prime``        ``if` `(prime(n ``-` `2``)):``            ``return` `2` `        ``# There exists 3 primes x1, x2, x3``        ``# such that x1 + x2 + x3 = n``        ``else``:``            ``return` `3` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `27` `    ``# Function call``    ``print``(Min_Sum(n))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program to Sum of all second largest``// divisors after splitting a number into one or more parts``using` `System;` `class` `GFG``{` `// Function to find a number is prime or not``static` `bool` `prime(``int` `n)``{``    ``if` `(n == 1)``        ``return` `false``;` `    ``// If there is any divisor``    ``for` `(``int` `i = 2; i * i <= n; ++i)``        ``if` `(n % i == 0)``            ``return` `false``;` `    ``return` `true``;``}` `// Function to find the sum of all second largest divisor``// after splitting a number into one or more parts``static` `int` `Min_Sum(``int` `n)``{``    ``// If number is prime``    ``if` `(prime(n))``        ``return` `1;` `    ``// If n is even``    ``if` `(n % 2 == 0)``        ``return` `2;` `    ``// If the number is odd``    ``else` `{` `        ``// If N-2 is prime``        ``if` `(prime(n - 2))``            ``return` `2;` `        ``// There exists 3 primes x1, x2, x3``        ``// such that x1 + x2 + x3 = n``        ``else``            ``return` `3;``    ``}``}` `// Driver code``public` `static` `void` `Main ()``{``    ``int` `n = 27;` `    ``// Function call``    ``Console.WriteLine( Min_Sum(n));``}``}` `// This code is contributed by anuj_6`

## Javascript

 ``
Output:

`3`

Time complexity: O(sqrt(N))

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up