Find count of digits in a number that divide the number

Given a positive integer n. The task is to find count of digits of number which evenly divides the number n.

Examples:

Input : n = 12
Output : 2
1 and 2 divide 12.

Input : n = 1012
Output : 3
1, 1 and 2 divide 1012.

The idea is to find each digit of the number n by modulus 10 and then check whether it divides n or not. Accordingly, increment the counter. Notice that the digit can be 0, so take care of that case.

Below is implementation of this approach:

C++



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// C++ program to count number of digits
// that divides the number.
#include <bits/stdc++.h>
using namespace std;
  
// Return the number of digits that divides
// the number.
int countDigit(int n)
{
    int temp = n, count = 0;
    while (temp != 0) {
        // Fetching each digit of the number
        int d = temp % 10;
        temp /= 10;
  
        // Checking if digit is greater than 0
        // and can divides n.
        if (d > 0 && n % d == 0)
            count++;
    }
  
    return count;
}
  
// Driven Program
int main()
{
    int n = 1012;
  
    cout << countDigit(n) << endl;
    return 0;
}

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Java

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// Java program to count number of digits
// that divides the number.
  
class Test {
    // Return the number of digits that divides
    // the number.
    static int countDigit(int n)
    {
        int temp = n, count = 0;
        while (temp != 0) {
            // Fetching each digit of the number
            int d = temp % 10;
            temp /= 10;
  
            // Checking if digit is greater than 0
            // and can divides n.
            if (d > 0 && n % d == 0)
                count++;
        }
  
        return count;
    }
  
    // Driver method
    public static void main(String args[])
    {
        int n = 1012;
        System.out.println(countDigit(n));
    }
}

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Python3

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# Python3 code to count number of 
# digits that divides the number.
  
# Return the number of digits
# that divides the number.
def countDigit (n):
    temp = n
    count = 0
    while temp != 0:
          
        # Fetching each digit 
        # of the number
        d = temp % 10
        temp /= 10
      
        # Checking if digit is greater
        # than 0 and can divides n.
        if d > 0 and n % d == 0:
            count += 1
    return count
      
# Driven Code
n = 1012
print(countDigit(n))
  
# This code is contributed by "Sharad_Bhardwaj".

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C#

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// C# program to count number of digits
// that divides the number.
using System;
  
class GFG {
  
    // Return the number of digits that
    // divides the number.
    static int countDigit(int n)
    {
        int temp = n, count = 0;
        while (temp != 0) {
  
            // Fetching each digit of
            // the number
            int d = temp % 10;
            temp /= 10;
  
            // Checking if digit is
            // greater than 0 and can
            // divides n.
            if (d > 0 && n % d == 0)
                count++;
        }
  
        return count;
    }
  
    // Driver method
    public static void Main()
    {
        int n = 1012;
  
        Console.Write(countDigit(n));
    }
}
  
// This code is contributed by parashar.

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PHP

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<?php
// PHP program to count 
// number of digits 
// that divides the number.
  
// Return the number of 
// digits that divides
// the number.
function countDigit($n)
{
    $temp = $n
    $count = 0;
      
    while ($temp != 0)
    {
          
        // Fetching each digit
        // of the number
        $d = $temp % 10;
        $temp /= 10;
          
        // Checking if digit 
        // is greater than 0
        // and can divides n.
        if ($d > 0 && $n % $d == 0)
        $count++;
    }
  
    return $count;
}
  
    // Driver Code
    $n = 1012;
    echo countDigit($n), "\n";
      
// This code is contributed by ajit
?>

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Output:

3

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : parashar, jit_t