Find largest sum of digits in all divisors of n
Last Updated :
19 Jul, 2025
Given an integer number n, find the 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
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 the largest sum of digits.
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;
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 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 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# 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 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.
?>
<script>
// Javascript program to find maximum
// sum of digits in all divisors
// of n numbers.
// Function to get sum of digits
function getSum(n)
{
let sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = Math.floor(n/10);
}
return sum;
}
// returns maximum sum
function largestDigitSumdivisior(n)
{
let res = 0;
for (let 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
let n = 14;
document.write(largestDigitSumdivisior(n)
+ "<br>");
// This code is contributed by Mayank Tyagi
</script>
Time Complexity: O(n*log10 (n))
Auxiliary Space: O(1)
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 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
# 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# 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 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.
?>
<script>
// JavaScript program to find
// maximum sum of digits
// in all divisors of n
// numbers.
// Function to get
// sum of digits
function getSum(n)
{
var sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = parseInt(n / 10);
}
return sum;
}
// returns maximum sum
function largestDigitSumdivisior(n)
{
var res = 0;
// traverse till sqrt(n)
for (var 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
var n = 14;
document.write(largestDigitSumdivisior(n));
</script>
Time Complexity: O(sqrt(n) log n)
Auxiliary Space: O(1) as it is using constant space for variables
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