# Find largest sum of digits in all divisors of n

Given a integer number n, find largest sum of digits in all divisors of n.

Examples :

```Input : n = 12
Output : 6
Explanation:
The divisors are: 1 2 3 4 6 12.
6 is maximum sum among all divisors

Input : n = 68
Output : 14
Explanation:
The divisors are: 1 2 4 68
68 consists of maximum sum of digit
```

## Recommended: Please solve it on PRACTICE first, before moving on to the solution.

Naive approach:
The idea is simple, we find all divisors of a number one by one. For every divisor, we compute sum of digits. Finally, we return largest sum of digits.

Time Complexity: O(n log n)

Below is the implementation of above approach:

## CPP

 `// CPP program to find maximum   ` `// sum of digits in all divisors ` `// of n numbers. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to get sum of digits ` `int` `getSum(``int` `n) ` `{  ` `int` `sum = 0; ` `while` `(n != 0) ` `{ ` `    ``sum = sum + n % 10; ` `    ``n = n/10; ` `} ` `return` `sum; ` `} ` ` `  `// returns maximum sum ` `int` `largestDigitSumdivisior(``int` `n) ` `{ ` `    ``int` `res = 0; ` `    ``for` `(``int` `i = 1; i <= n; i++)  ` ` `  `        ``// if i is factor of n  ` `        ``// then push the divisor  ` `        ``// in the stack. ` `        ``if` `(n % i == 0)  ` `        ``res = max(res, getSum(i)); ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 14; ` `    ``cout << largestDigitSumdivisior(n) ` `         ``<< endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find maximum ` `// sum of digits in all divisors ` `// of n numbers. ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GfG ` `{ ` `     `  `    ``// Function to get  ` `    ``// sum of digits ` `    ``public` `static` `int` `getSum(``int` `n) ` `    ``{  ` `        ``int` `sum = ``0``; ` `        ``while` `(n != ``0``) ` `        ``{ ` `            ``sum = sum + n % ``10``; ` `            ``n = n/``10``; ` `        ``} ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// returns maximum sum ` `    ``public` `static` `int` `largestDigitSumdivisior(``int` `n) ` `    ``{ ` `        ``int` `res = ``0``; ` `        ``for` `(``int` `i = ``1``; i <= n; i++)  ` ` `  `            ``// if i is factor of n   ` `            ``// then push the divisor  ` `            ``// in the stack. ` `            ``if` `(n % i == ``0``)  ` `            ``res = Math.max(res, getSum(i)); ` ` `  `        ``return` `res; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String argc[]){ ` `        ``int` `n = ``14``; ` `         `  `        ``System.out.println(largestDigitSumdivisior(n)); ` `    ``} ` `     `  `} ` `// This code is contributed ` `// by Sagar Shukla `

## Python3

 `# Python3 code to find  ` `# maximum sum of digits  ` `# in all divisors of n numbers. ` ` `  `# Function to get sum of digits ` `def` `getSum( n ): ` `    ``sum` `=` `0` `    ``while` `n !``=` `0``: ` `        ``sum` `=` `sum` `+` `n ``%` `10` `        ``n ``=` `int``( n ``/` `10` `) ` `    ``return` `sum` ` `  `# returns maximum sum ` `def` `largestDigitSumdivisior( n ): ` `    ``res ``=` `0` `    ``for` `i ``in` `range``(``1``, n ``+` `1``): ` ` `  `        ``# if i is factor of n  ` `        ``# then push the divisor ` `        ``# in the stack. ` `        ``if` `n ``%` `i ``=``=` `0``: ` `            ``res ``=` `max``(res, getSum(i)) ` ` `  `    ``return` `res ` ` `  ` `  `# Driver Code ` `n ``=` `14` `print``(largestDigitSumdivisior(n) ) ` ` `  `# This code is contributed ` `# by "Sharad_Bhardwaj". `

## C#

 `// C# program to find maximum  ` `// sum of digits in all  ` `// divisors of n numbers. ` `using` `System; ` ` `  `class` `GfG  ` `{ ` `     `  `    ``// Function to get ` `    ``// sum of digits ` `    ``public` `static` `int` `getSum(``int` `n) ` `    ``{  ` `        ``int` `sum = 0; ` `        ``while` `(n != 0) ` `        ``{ ` `            ``sum = sum + n % 10; ` `            ``n = n / 10; ` `        ``} ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// returns maximum sum ` `    ``public` `static` `int` `largestDigitSumdivisior(``int` `n) ` `    ``{ ` `        ``int` `res = 0; ` `        ``for` `(``int` `i = 1; i <= n; i++)  ` ` `  `            ``// if i is factor of n  ` `            ``// then push the divisor ` `            ``// in the stack. ` `            ``if` `(n % i == 0)  ` `            ``res = Math.Max(res, getSum(i)); ` ` `  `        ``return` `res; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 14; ` `         `  `        ``Console.WriteLine(largestDigitSumdivisior(n)); ` `    ``} ` `     `  `} ` ` `  `// This code is contributed by vt_m `

## PHP

 ` `

Output :

```7
```

An efficient approach will be to find the divisors in O(sqrt n). We follow the same steps as above , just iterate till sqrt(n) and get i and n/i as their divisors whenever n%i==0.

Below is the implementation of the above approach:

## CPP

 `// CPP program to find ` `// maximum sum of digits ` `// in all divisors of n  ` `// numbers. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to get ` `// sum of digits  ` `int` `getSum(``int` `n) ` `{  ` `int` `sum = 0; ` `while` `(n != 0) ` `{ ` `    ``sum = sum + n % 10; ` `    ``n = n / 10; ` `} ` `return` `sum; ` `} ` ` `  `// returns maximum sum ` `int` `largestDigitSumdivisior(``int` `n) ` `{ ` `    ``int` `res = 0; ` `     `  `    ``// traverse till sqrt(n)  ` `    ``for` `(``int` `i = 1; i <= ``sqrt``(n); i++)  ` ` `  `        ``// if i is factor of ` `        ``// n then push the ` `        ``// divisor in the stack. ` `        ``if` `(n % i == 0) ` `        ``{ ` `            ``// check for both the divisors ` `            ``res = max(res, getSum(i)); ` `            ``res = max(res,getSum(n / i)); ` `        ``}      ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `n = 14; ` `    ``cout << largestDigitSumdivisior(n)  ` `         ``<< endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find maximum  ` `// sum of digits in all divisors ` `// of n numbers. ` ` `  `import` `java.io.*; ` `import` `java.math.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to get  ` `    ``// sum of digits  ` `    ``static` `int` `getSum(``int` `n) ` `    ``{ ` `        ``int` `sum = ``0``; ` `        ``while` `(n != ``0``) ` `        ``{ ` `            ``sum = sum + n % ``10``; ` `            ``n = n / ``10``; ` `        ``} ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// returns maximum sum ` `    ``static` `int` `largestDigitSumdivisior(``int` `n) ` `    ``{ ` `        ``int` `res = ``0``; ` ` `  `        ``// traverse till sqrt(n) ` `        ``for` `(``int` `i = ``1``; i <= Math.sqrt(n); i++) ` `        ``{ ` ` `  `            ``// if i is factor of ` `            ``// n then push the ` `            ``// divisor in the stack. ` `            ``if` `(n % i == ``0``)  ` `            ``{ ` `                 `  `                ``// check for both the divisors ` `                ``res = Math.max(res, getSum(i)); ` `                ``res = Math.max(res, getSum(n / i)); ` `            ``} ` ` `  `        ``} ` `         `  `        ``return` `res; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``14``; ` `        ``System.out.println(largestDigitSumdivisior(n)); ` `    ``} ` `} ` ` `  `// This code is contributed ` `// by Nikita Tiwari `

## Python3

 `# Python 3 program ` `# to find maximum ` `# sum of digits in  ` `# all divisors of ` `# n numbers ` `import` `math ` ` `  `# Function to get  ` `# sum of digits  ` `def` `getSum(n) : ` `    ``sm ``=` `0` `    ``while` `(n !``=` `0``) : ` `        ``sm ``=` `sm ``+` `n ``%` `10` `        ``n ``=` `n ``/``/` `10` `         `  `    ``return` `sm ` `     `  `     `  `# returns maximum sum ` `def` `largestDigitSumdivisior(n) : ` `    ``res ``=` `0` `     `  `    ``# traverse till sqrt(n)  ` `    ``for` `i ``in` `range``(``1``, (``int``)(math.sqrt(n))``+``1``) : ` `         `  `        ``# if i is factor of n then  ` `        ``# push the divisor in the ` `        ``# stack. ` `        ``if` `(n ``%` `i ``=``=` `0``) : ` ` `  `            ``# check for both the ` `            ``# divisors ` `            ``res ``=` `max``(res, getSum(i)) ` `            ``res ``=` `max``(res, getSum(n ``/``/` `i)) ` `             `  `    ``return` `res ` ` `  `# Driver Code ` `n ``=` `14` `print``(largestDigitSumdivisior(n)) ` ` `  `#This code is contributed  ` `# by Nikita Tiwari `

## C#

 `// C# program to find maximum sum ` `// of digits in all divisors of n  ` `// numbers. ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to get  ` `    ``// sum of digits  ` `    ``static` `int` `getSum(``int` `n) ` `    ``{ ` `        ``int` `sum = 0; ` `         `  `        ``while` `(n != 0) ` `        ``{ ` `            ``sum = sum + n % 10; ` `            ``n = n / 10; ` `        ``} ` `         `  `        ``return` `sum; ` `    ``} ` ` `  `    ``// returns maximum sum ` `    ``static` `int` `largestDigitSumdivisior(``int` `n) ` `    ``{ ` `        ``int` `res = 0; ` ` `  `        ``// traverse till sqrt(n) ` `        ``for` `(``int` `i = 1; i <= Math.Sqrt(n); i++) ` `        ``{ ` ` `  `            ``// if i is factor of n then push the ` `            ``// divisor in the stack. ` `            ``if` `(n % i == 0)  ` `            ``{ ` `                 `  `                ``// check for both the divisors ` `                ``res = Math.Max(res, getSum(i)); ` `                ``res = Math.Max(res, getSum(n / i)); ` `            ``} ` ` `  `        ``} ` `         `  `        ``return` `res; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 14; ` `         `  `        ``Console.WriteLine(largestDigitSumdivisior(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Vt_m `

## PHP

 ` `

Output :

```7
```

Time Complexity: O(sqrt(n) log n)

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