# Maximum sum after repeatedly dividing N by a divisor

Given an integer N. The task is to find the maximum possible sum of intermediate values (Including N and 1) attained after applying the beow operation:

Divide N by any divisor (>1) until it becomes 1.

Examples:

```Input: N = 10
Output: 16
Initially, N=10
1st Division -> N = 10/2 = 5
2nd Division -> N= 5/5 = 1

Input: N = 8
Output: 15
Initially, N=8
1st Division -> N = 8/2 = 4
2nd Division -> N= 4/2 = 2
3rd Division -> N= 2/2 = 1
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Since the task is to maximize the sum of values after each step, try to maximize individual values. So, reduce the value of N by as little as possible. To achieve that, we divide N by its smallest divisor.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the smallest divisor ` `int` `smallestDivisor(``int` `n) ` `{ ` `    ``int` `mx = ``sqrt``(n); ` `    ``for` `(``int` `i = 2; i <= mx; i++) ` `        ``if` `(n % i == 0) ` `            ``return` `i; ` `    ``return` `n; ` `} ` ` `  `// Function to find the maximum sum ` `int` `maxSum(``int` `n) ` `{ ` `    ``long` `long` `res = n; ` `    ``while` `(n > 1) { ` `        ``int` `divi = smallestDivisor(n); ` `        ``n /= divi; ` `        ``res += n; ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 34; ` `    ``cout << maxSum(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to find the smallest divisor ` `static` `double` `smallestDivisor(``int` `n) ` `{ ` `    ``double` `mx = Math.sqrt(n); ` `    ``for` `(``int` `i = ``2``; i <= mx; i++) ` `        ``if` `(n % i == ``0``) ` `            ``return` `i; ` `    ``return` `n; ` `} ` ` `  `// Function to find the maximum sum ` `static` `double` `maxSum(``int` `n) ` `{ ` `    ``long` `res = n; ` `    ``while` `(n > ``1``)  ` `    ``{ ` `        ``double` `divi = smallestDivisor(n); ` `        ``n /= divi; ` `        ``res += n; ` `    ``} ` `    ``return` `res; ` `} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `n = ``34``; ` `        ``System.out.println (maxSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by jit_t. `

## Python3

 `from` `math ``import` `sqrt ` `# Python 3 implementation of the above approach ` ` `  `# Function to find the smallest divisor ` `def` `smallestDivisor(n): ` `    ``mx ``=` `int``(sqrt(n)) ` `    ``for` `i ``in` `range``(``2``, mx ``+` `1``, ``1``): ` `        ``if` `(n ``%` `i ``=``=` `0``): ` `            ``return` `i ` `    ``return` `n ` ` `  `# Function to find the maximum sum ` `def` `maxSum(n): ` `    ``res ``=` `n ` `    ``while` `(n > ``1``): ` `        ``divi ``=` `smallestDivisor(n) ` `        ``n ``=` `int``(n``/``divi) ` `        ``res ``+``=` `n ` `     `  `    ``return` `res ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `34` `    ``print``(maxSum(n)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function to find the smallest divisor  ` `    ``static` `double` `smallestDivisor(``int` `n)  ` `    ``{  ` `        ``double` `mx = Math.Sqrt(n);  ` `        ``for` `(``int` `i = 2; i <= mx; i++)  ` `            ``if` `(n % i == 0)  ` `                ``return` `i;  ` `        ``return` `n;  ` `    ``}  ` `     `  `    ``// Function to find the maximum sum  ` `    ``static` `double` `maxSum(``int` `n)  ` `    ``{  ` `        ``long` `res = n;  ` `        ``while` `(n > 1)  ` `        ``{  ` `            ``double` `divi = smallestDivisor(n);  ` `            ``n /= (``int``)divi;  ` `            ``res += n;  ` `        ``}  ` `        ``return` `res;  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `n = 34;  ` `        ``Console.WriteLine(maxSum(n));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga.  `

## PHP

 ` 1) ` `    ``{ ` `        ``\$divi` `= smallestDivisor(``\$n``); ` `        ``\$n` `/= ``\$divi``; ` `        ``\$res` `+= ``\$n``; ` `    ``} ` `    ``return` `\$res``; ` `} ` ` `  `    ``// Driver Code ` `    ``\$n` `= 34; ` `    ``echo` `maxSum(``\$n``); ` ` `  `#This code is contributed by akt_mit. ` `?> `

Output:

```52
```

Time Complexity: O(sqrt(n)*log(n)) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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