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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 <bits/stdc++.h>
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




<script>
// Javascript program to find sum of all second largest divisor
// after splitting a number into one or more parts
 
// Function to find a number is prime or not
function prime(n)
{
    if (n == 1)
        return false;
 
    // If there is any divisor
    for (let 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
function 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
    let n = 27;
 
    // Function call
    document.write(Min_Sum(n));
 
// This code is contributed by Mayank Tyagi
 
</script>
Output: 



3

 

Time complexity: O(sqrt(N))
 

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