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

Given an integer N where . The task is to check whether the number is not divisible by any of its digit. If the given number N is divisible by any of its digits then print “YES” else print “NO”.

Examples:

Input : N = 5115
Output : YES
Explanation: 5115 is divisible by both 1 and 5.
So print YES.

Input : 27
Output : NO
Explanation: 27 is not divisible by 2 or 7


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach : The idea to solve the problem is to extract the digits of the number one by one and check if the number is divisible by any of its digit. If it is divisible by any of it’s digit then print YES otherwise print NO.

Below is the implementation of above approach:

## C++

 // CPP implementation of above approach  #include  using namespace std;     // Function to check if given number is divisible  // by any of its digits  string isDivisible(long long int n)  {      long long int temp = n;         // check if any of digit divides n      while (n) {          int k = n % 10;             // check if K divides N          if (temp % k == 0)              return "YES";             n /= 10;      }         return "NO";  }     // Driver Code  int main()  {      long long int n = 9876543;         cout << isDivisible(n);         return 0;  }

## Java

 // Java implementation of above approach     class GFG   {         // Function to check if given number is divisible      // by any of its digits      static String isDivisible(int n)       {          int temp = n;             // check if any of digit divides n          while (n > 0)          {              int k = n % 10;                 // check if K divides N              if (temp % k == 0)              {                  return "YES";              }              n /= 10;          }             return "NO";      }         // Driver Code      public static void main(String[] args)       {          int n = 9876543;          System.out.println(isDivisible(n));      }  }      // This code is contributed by 29AjayKumar

## Python3

 # Python program implementation of above approach     # Function to check if given number is   # divisible by any of its digits  def isDivisible(n):      temp = n         # check if any of digit divides n      while(n):          k = n % 10            # check if K divides N          if(temp % k == 0):              return "YES"            n /= 10;         # Number is not divisible by       # any of digits      return "NO"    # Driver Code  n = 9876543 print(isDivisible(n))     # This code is contributed by  # Sanjit_Prasad

## C#

 // C# implementation of above approach  using System;     class GFG   {         // Function to check if given number is divisible      // by any of its digits      static String isDivisible(int n)       {          int temp = n;             // check if any of digit divides n          while (n > 0)          {              int k = n % 10;                 // check if K divides N              if (temp % k == 0)              {                  return "YES";              }              n /= 10;          }             return "NO";      }         // Driver Code      public static void Main(String[] args)       {          int n = 9876543;          Console.WriteLine(isDivisible(n));      }  }     // This code is contributed by PrinciRaj1992

## PHP

 

Output:

YES


Time Complexity: O(log(N))
Auxiliary Space: O(1)

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