Program to check if a number is divisible by sum of its digits

Given an integer N, the task is to check whether the number is divisible by the sum of its digits or not. If divisible, then print “YES” else print “NO”.

Examples:

Input: N = 12 
Output: YES
Explanation: 
    As sum of digits of 12 = 1 + 2 = 3
    and 12 is divisible by 3 
    So the output is YES

Input: N = 123
Output: NO

Approach: The idea to solve the problem is to extract the digits of the number and add them. Then check if the number is divisible by the sum of its digit. If it is divisible then print YES otherwise print NO.



Below is the implementation of above approach:

Implementation:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP implementation of above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to check
// if the given number is divisible
// by sum of its digits
string isDivisible(long long int n)
{
    long long int temp = n;
  
    // Find sum of digits
    int sum = 0;
    while (n) {
        int k = n % 10;
        sum += k;
        n /= 10;
    }
  
    // check if sum of digits divides n
    if (temp % sum == 0)
        return "YES";
  
    return "NO";
}
  
// Driver Code
int main()
{
    long long int n = 123;
  
    cout << isDivisible(n);
  
    return 0;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java implementation of above approach  


class GFG 


{ 


  


    // Function to check if the  


    // given number is divisible  


    // by sum of its digits  


    static String isDivisible(long n)  


    {  


        long temp = n;  


      


        // Find sum of digits  


        int sum = 0;  


        while (n != 0)  


        {  


            int k = (int) n % 10;  


            sum += k;  


            n /= 10;  


        }  


      


        // check if sum of digits divides n  


        if (temp % sum == 0)  


            return "YES";  


      


        return "NO";  


    }  


      


    // Driver Code  


    public static void main(String []args) 


    {  


        long n = 123;  


        System.out.println(isDivisible(n));  


    }  


} 


  


// This code is contributed by Ryuga 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python 3 implementation of above approach 


  


# Function to check if the given number  


# is divisible by sum of its digits 


def isDivisible(n): 


  


    temp = n 


  


    # Find sum of digits 


    sum = 0; 


    while (n):  


      


        k = n % 10; 


        sum += k; 


        n /= 10; 


      


    # check if sum of digits divides n 


    if (temp % sum == 0): 


        return "YES"; 


  


    return "NO"; 


  


# Driver Code 


n = 123; 


  


print(isDivisible(n)); 


  


# This code is contributed by  


# Akanksha Rai 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# implementation of above approach  


using System; 


  


class GFG 


{ 


  


    // Function to check if the  


    // given number is divisible  


    // by sum of its digits  


    static String isDivisible(long n)  


    {  


        long temp = n;  


      


        // Find sum of digits  


        int sum = 0;  


        while (n != 0)  


        {  


            int k = (int) n % 10;  


            sum += k;  


            n /= 10;  


        }  


      


        // check if sum of digits divides n  


        if (temp % sum == 0)  


            return "YES";  


      


        return "NO";  


    }  


      


    // Driver Code  


    public static void Main() 


    {  


        long n = 123;  


        Console.WriteLine(isDivisible(n));  


    }  


} 


  


// This code is contributed by anuj_67.. 



















                        chevron_right







                        filter_none









PHP














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















<?php 


// PHP implementation of above approach 


  


// Function to check if the given number  


// is divisible by sum of its digits 


function isDivisible($n) 


{ 


    $temp = $n; 


  


    // Find sum of digits 


    $sum = 0; 


    while ($n)  


    { 


        $k = $n % 10; 


        $sum += $k; 


        $n = (int)($n / 10); 


    } 


  


    // check if sum of digits divides n 


    if ($temp % $sum == 0) 


        return "YES"; 


  


    return "NO"; 


} 


  


// Driver Code 


$n = 123; 


print(isDivisible($n)); 


  


// This code is contributed  


// by chandan_jnu 


?> 



















                        chevron_right







                        filter_none











Output:


NO







GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details 





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
sahilshelangia
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.