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
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 = 9876543print(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 ?> |
YES
Time Complexity: O(log(N))
Auxiliary Space: O(1)
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