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++

// C++ 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;
}

Java

// 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

Python3

# 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

C#

// 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..

PHP

<?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
?>

Javascript

<script>

// Javascript 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 = parseInt(n / 10);
    }

    // Check if sum of digits divides n
    if (temp % sum == 0)
        return "YES";

    return "NO";
}

// Driver Code
let n = 123;

document.write(isDivisible(n));

// This code is contributed by sravan kumar

</script>

Output:

NO

Time Complexity: O(log₁₀n)

Auxiliary Space: O(1)

Method #2: Using string:

• Convert the given number to a string by taking a new variable.
• Traverse the string, Convert each element to integer and add this to sum.
• If the number is divisible by sum then print Yes
• Else print No

Below is the implementation of above approach:

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

string getResult(long long int n)
{

    // Converting integer to String
    string st = std::to_string(n);

    // Initialising sum to 0
    int sum = 0;

    // Traversing through the String
    for(char i : st)
    {

        // Converting character to int
        sum = sum + (int) i;
    }

    // Comparing number and sum
    if (n % sum == 0)
        return "Yes";
    else
        return "No";
}

// Driver Code
int main()
{
    int n = 123;

    // Passing this number to get result function
    cout << getResult(n);
    return 0;
}

// This code is contributed by 29AjayKumar

Java

// Java implementation of above approach
class GFG{

static String getResult(int n)
{

    // Converting integer to String
    String st = String.valueOf(n);

    // Initialising sum to 0
    int sum = 0;

    // Traversing through the String
    for(char i : st.toCharArray())
    {

        // Converting character to int
        sum = sum + (int) i;
    }

    // Comparing number and sum
    if (n % sum == 0)
        return "Yes";
    else
        return "No";
}

// Driver Code
public static void main(String[] args)
{
    int n = 123;

    // Passing this number to get result function
    System.out.println(getResult(n));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python implementation of above approach
def getResult(n):

    # Converting integer to string
    st = str(n)

    # Initialising sum to 0
    sum = 0
    length = len(st)

    # Traversing through the string
    for i in st:

        # Converting character to int
        sum = sum + int(i)

    # Comparing number and sum
    if (n % sum == 0):
        return "Yes"
    else:
        return "No"


# Driver Code
n = 123
# passing this number to get result function
print(getResult(n))

# this code is contributed by vikkycirus

C#

// C# implementation of above approach
using System;
public class GFG{

static String getResult(int n)
{

    // Converting integer to String
    String st = String.Join("",n);

    // Initialising sum to 0
    int sum = 0;

    // Traversing through the String
    foreach(char i in st.ToCharArray())
    {

        // Converting character to int
        sum = sum + (int) i;
    }

    // Comparing number and sum
    if (n % sum == 0)
        return "Yes";
    else
        return "No";
}

// Driver Code
public static void Main(String[] args)
{
    int n = 123;

    // Passing this number to get result function
    Console.WriteLine(getResult(n));
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// JavaScript implementation of above approach

function getResult(n)
{
    // Converting integer to String
    let st = (n).toString();

    // Initialising sum to 0
    let sum = 0;

    // Traversing through the String
    for(let i of st.split(""))
    {

        // Converting character to int
        sum = sum + parseInt(i);
    }

    // Comparing number and sum
    if (n % sum == 0)
        return "Yes";
    else
        return "No";
}

// Driver Code
let n = 123;

// Passing this number to get result function
document.write(getResult(n)+"<br>");

// This code is contributed by unknown2108

</script>

Output:

No

Time Complexity: O(N)