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# 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 = 12Output : YESSum of digits = 1+2 =3 and 3 divides 12.So, print YES.Input : N = 15Output : 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 ``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

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## PHP

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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 ``#include ``#include ``#include ` `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

## C++

 `#include ``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.