Skip to content
Related Articles

Related Articles

Program to check if a number is divisible by any of its digits
  • Last Updated : 25 Jan, 2019

Given an integer N where 1 \leq n \leq 10^{18}. 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

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 <bits/stdc++.h>
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




<?php
// PHP implementation of above approach 
  
// Function to check if given number 
// is divisible by any of its digits 
function 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"
  
        $n = floor($n / 10); 
    
  
    return "NO"
  
// Driver Code 
$n = 9876543; 
  
echo isDivisible($n); 
  
// This code is contributed by Ryuga
?>


Output:

YES

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :