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 <bits/stdc++.h> 
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

<?php 
// PHP program to find maximum 
// sum of digits in all 
// divisors of n numbers. 
  
// Function to get  
// sum of digits  
function getSum( $n) 
{  
    $sum = 0; 
    while ($n != 0) 
{ 
    $sum = $sum + $n % 10; 
    $n = $n/10; 
} 
return $sum; 
} 
  
// returns maximum sum 
function largestDigitSumdivisior( $n) 
{ 
    $res = 0; 
    for ($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 
    $n = 14; 
    echo largestDigitSumdivisior($n); 
  
// This code is contributed by anuj_67. 
?> 

Output :

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 <bits/stdc++.h> 
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

<?php 
// PHP program to find maximum 
// sum of digits in all 
// divisors of n numbers 
  
// Function to get 
// sum of digits  
function getSum($n) 
{  
    $sum = 0; 
while ($n != 0) 
{ 
    $sum = $sum + $n % 10; 
    $n = $n / 10; 
} 
return $sum; 
} 
  
// returns maximum sum 
function largestDigitSumdivisior( $n) 
{ 
    $res = 0; 
      
    // traverse till sqrt(n)  
    for ($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 
$n = 14; 
echo largestDigitSumdivisior($n); 
  
// This code is contributed by anuj_67. 
?> 

Output :

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

My Personal Notes

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

CPP

Java

Python3

C#

PHP

CPP

Java

Python3

C#

PHP

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