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++ 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 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 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# 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 |
<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 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 ?> |
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.
#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;
} |
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
}
} |
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 |
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
}
} |
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
|
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 –
#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;
} |
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" );
}
}
} |
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" )
|
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" );
}
}
} |
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" );
} |
YES NO
Time complexity: O(logn), where n is the number.
Auxiliary space: O(logn), due to the recursive call.