Check if the sum of digits of a number N divides it

Given a number N. The task is to check if the sum of digits of the given number divides the number or not. If it divides it then print YES otherwise print NO.
Examples: 

Input : N = 12
Output : YES
Sum of digits = 1+2 =3 and 3 divides 12.
So, print YES.
Input : N = 15
Output : NO

Extract digits of the number and calculate the sum of all of its digits and check if the sum of digits dives N or not.
Below is the implementation of the above approach: 

C++

// C++ program to check if sum of
// digits of a number divides it

#include <iostream>
using namespace std;

// Function to check if sum of
// digits of a number divides it
int isSumDivides(int N)
{
    int temp = N;

    int sum = 0;

    // Calculate sum of all of digits of N
    while (temp) {
        sum += temp % 10;
        temp /= 10;
    }

    if (N % sum == 0)
        return 1;
    else
        return 0;
}

// Driver Code
int main()
{
    int N = 12;

    if (isSumDivides(N))
        cout << "YES";
    else
        cout << "NO";

    return 0;
}

Java

// Java program to check if sum of
// digits of a number divides it

import java.util.*;
import java.lang.*;

class GFG
{
// Function to check if sum of
// digits of a number divides it
static int isSumDivides(int N)
{
    int temp = N;

    int sum = 0;

    // Calculate sum of all of digits of N
    while (temp > 0)
    {
        sum += temp % 10;
        temp /= 10;
    }

    if (N % sum == 0)
        return 1;
    else
        return 0;
}

// Driver Code
public static void main(String args[])
{
    int N = 12;

    if (isSumDivides(N) == 1)
        System.out.print("YES");
    else
        System.out.print("NO");
}
}

// This code is contributed
// by Akanksha Rai(Abby_akku)

Python3

# Python3 program to check if sum of
# digits of a number divides it

# Function to check if sum of
# digits of a number divides it
def isSumDivides(N):

    temp = N

    sum = 0

    # Calculate sum of all of
    # digits of N
    while (temp):
        sum += temp % 10
        temp = int(temp / 10)

    if (N % sum == 0):
        return 1
    else:
        return 0

# Driver Code
if __name__=='__main__':
    N = 12

    if (isSumDivides(N)):
        print("YES")
    else:
        print("NO")

# This code is contributed by
# mits

C#

// C# program to check if sum of
// digits of a number divides it
using System;

// Function to check if sum of
// digits of a number divides it
class GFG
{
public int isSumDivides(int N)
{
    int temp = N, sum = 0;

    // Calculate sum of all of
    // digits of N
    while (temp > 0)
    {
        sum += temp % 10;
        temp /= 10;
    }

    if (N % sum == 0)
        return 1;
    else
        return 0;
}

// Driver Code
public static void Main()
{
    GFG g = new GFG();
    int N = 12;

    if (g.isSumDivides(N) > 0)
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}

// This code is contributed by Soumik

Javascript

<script>
// javascript program to check if sum of
// digits of a number divides it
// Function to check if sum of
    // digits of a number divides it
    function isSumDivides(N) {
        var temp = N;

        var sum = 0;

        // Calculate sum of all of digits of N
        while (temp > 0) {
            sum += temp % 10;
            temp = parseInt(temp/10);
        }

        if (N % sum == 0)
            return 1;
        else
            return 0;
    }

    // Driver Code

        var N = 12;

        if (isSumDivides(N) == 1)
            document.write("YES");
        else
            document.write("NO");

// This code contributed by aashish1995
</script>

PHP

<?php
// PHP program to check if sum of
// digits of a number divides it

// Function to check if sum of
// digits of a number divides it
function isSumDivides($N)
{
    $temp = $N;

    $sum = 0;

    // Calculate sum of all of
    // digits of N
    while ($temp)
    {
        $sum += $temp % 10;
        $temp = (int)$temp / 10;
    }

    if ($N % $sum == 0)
        return 1;
    else
        return 0;
}

// Driver Code
$N = 12;

if (isSumDivides($N))
    echo "YES";
else
    echo "NO";

// This code is contributed by ajit
?>

Output

YES

Time Complexity : O(logn) 
Auxiliary Space: O(1), since no extra space has been taken.

Method:  Using the map function, split method, and the sum function:

This approach first converts the number N to a string and then splits it into a list of individual digits using the map function and the split method. It then calculates the sum of the digits using the sum function and checks if the sum divides the number N.

C++

#include <iostream>
#include <string>
#include <vector>
#include <numeric>

using namespace std;

bool isSumDivides(int N) {
    // Convert the number to a string and split it into individual digits
    string str_N = to_string(N);
    vector<int> digits;
    for (char c : str_N) {
        digits.push_back(c - '0');
    }

    // Calculate the sum of the digits
    int sum_of_digits = accumulate(digits.begin(), digits.end(), 0);

    // Check if the sum of the digits divides the number
    return N % sum_of_digits == 0;
}

int main() {
    int N = 12;
    cout << boolalpha << isSumDivides(N) << endl; // should print true

    N = 15;
    cout << boolalpha << isSumDivides(N) << endl; // should print false

    return 0;
}

Java

import java.util.Arrays;

public class Main {
    public static boolean isSumDivides(int N) {
        // Convert the number to a string and split it into individual digits
        int[] digits = Arrays.stream(Integer.toString(N).split("")).mapToInt(Integer::parseInt).toArray();

        // Calculate the sum of the digits
        int sum_of_digits = Arrays.stream(digits).sum();

        // Check if the sum of the digits divides the number
        return N % sum_of_digits == 0;
    }

    public static void main(String[] args) {
        int N = 12;
        System.out.println(isSumDivides(N)); // should print true

        N = 15;
        System.out.println(isSumDivides(N)); // should print false
    }
}

Python3

def isSumDivides(N):
    # Convert the number to a string and split it into individual digits
    digits = list(map(int, str(N)))

    # Calculate the sum of the digits
    sum_of_digits = sum(digits)

    # Check if the sum of the digits divides the number
    return N % sum_of_digits == 0

# Test the function
N = 12
print(isSumDivides(N)) # should print True

N = 15
print(isSumDivides(N)) # should print False
#This code is contributed by Edula Vinay Kumar Reddy

C#

using System;
using System.Linq;

public class GFG {
    public static bool IsSumDivides(int N) {
        // Convert the number to a string and split it into individual digits
        int[] digits = Array.ConvertAll(N.ToString().ToCharArray(), c => Convert.ToInt32(c.ToString()));

        // Calculate the sum of the digits
        int sum_of_digits = digits.Sum();

        // Check if the sum of the digits divides the number
        return N % sum_of_digits == 0;
    }

    public static void Main() {
        int N = 12;
        Console.WriteLine(IsSumDivides(N)); // should print true

        N = 15;
        Console.WriteLine(IsSumDivides(N)); // should print false
    }
}

Javascript

function isSumDivides(N)
{

  // Convert the number to a string and split it into individual digits
  let digits = N.toString().split("").map(Number);

  // Calculate the sum of the digits
  let sum_of_digits = digits.reduce((acc, cur) => acc + cur, 0);

  // Check if the sum of the digits divides the number
  return N % sum_of_digits == 0;
}

let N = 12;
console.log(isSumDivides(N)); // should print true

N = 15;
console.log(isSumDivides(N)); // should print false

Output

True
False

Time complexity: O(n), where n is the number of digits in N
Auxiliary space: O(n), since a list of size n is created to store the digits of N

Method: using Recursion –

In this approach, we will use recursion to find out the sum of all the digits in a number and then divide that number with the sum we got.

Stepwise Explanation – 

Step – 1: We will use a function in which we will calculate and return the sum of the digits of a number which we will pass as argument.

Step – 2:  Outside of the function we will use the result we got from the function and divide it with the number we took first and check if the remainder returned is 0 or not.

If 0 then it is entirely divisible otherwise not.

Example – 

Let the number we took be 12.

Step – 1 : 12 % 10 is 2 + send (12/10) in next step

Step – 2 – 1 % 10 is 1 + send (1/10) in next step

Step – 3 : No more digits left 

so from first step we got 2

from second step we got 1

therefore the sum is – 2+1 = 3

and 12 % 3 == 0.

So our code will return True

Below is the Implementation –

C++

#include <bits/stdc++.h>
using namespace std;

// Function to check sum of digit using recursion
int sum_of_digits(int n)
{
    if (n == 0)
    return 0;
    return (n % 10 + sum_of_digits(n / 10));
}

// Driven code
int main()
{
    // Example 1
    int num_1 = 12;

      // Example 2
      int num_2 = 15;

    int result_1 = num_1 % sum_of_digits(num_1);
      int result_2 = num_2 % sum_of_digits(num_2);

      // First Result
    if (result_1 == 0){
      cout << "YES" << endl;
    }
      else{
      cout << "NO" << endl;
    }

      // Second Result
      if (result_2 == 0){
      cout << "YES";
    }
      else{
      cout << "NO";
    }



    return 0;
}

Java

import java.util.*;

class GFG {
    // Function to check sum of digits using recursion
    static int sumOfDigits(int n) {
        if (n == 0)
            return 0;
        return (n % 10 + sumOfDigits(n / 10));
    }

    // Driver code
    public static void main(String[] args) {
        // Example 1
        int num1 = 12;

        // Example 2
        int num2 = 15;

        int result1 = num1 % sumOfDigits(num1);
        int result2 = num2 % sumOfDigits(num2);

        // First Result
        if (result1 == 0) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }

        // Second Result
        if (result2 == 0) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}

Python3

def sum_of_digits(n):
    if n == 0:
        return 0
    return (n % 10 + sum_of_digits(int(n / 10)))


# Example - 1
num_1 = 12
result_1 = num_1 % sum_of_digits(num_1)

# Example - 2
num_2 = 15
result_2 = num_2 % sum_of_digits(num_2)


print("YES" if result_1 == 0 else "NO")
print("YES" if result_2 == 0 else "NO")

C#

using System;

class GFG
{
    // Function to check sum of digit using recursion
    static int SumOfDigits(int n)
    {
        if (n == 0)
            return 0;
        return (n % 10 + SumOfDigits(n / 10));
    }

    static void Main(string[] args)
    {
        // Example 1
        int num_1 = 12;

        // Example 2
        int num_2 = 15;

        int result_1 = num_1 % SumOfDigits(num_1);
        int result_2 = num_2 % SumOfDigits(num_2);

        // First Result
        if (result_1 == 0)
        {
            Console.WriteLine("YES");
        }
        else
        {
            Console.WriteLine("NO");
        }

        // Second Result
        if (result_2 == 0)
        {
            Console.WriteLine("YES");
        }
        else
        {
            Console.WriteLine("NO");
        }
    }
}

Javascript

function sumOfDigits(n){
    if (n == 0){
        return 0;
    }
    return (n%10+sumOfDigits(parseInt(n/10)));
}

let num_1 = 12;
let num_2 = 15;

let result_1 = num_1 % sumOfDigits(num_1);
let result_2 = num_2 % sumOfDigits(num_2);

if (result_1 == 0){
    console.log("YES \n");
}
else{
    console.log("NO \n");
}

if (result_2 == 0){
    console.log("YES \n");
}
else{
    console.log("NO \n");
}

Output

YES
NO

Time complexity: O(logn), where n is the number.
Auxiliary space: O(logn), due to the recursive call.