# Count of numbers from the range [L, R] which contains at least one digit that divides K

Given three positive integers L, R and K.The task is to find the count of all the numbers from the range [L, R] that contains at least one digit which divides the number K.

Examples:

Input: L = 5, R = 11, K = 10
Output: 3
5, 10 and 11 are only such numbers.

Input: L = 32, R = 38, K = 13
Output: 0

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

Approach: Initialise count = 0 and for every element in the range [L, R], check if it contains at least one digit that divides K. If yes then increment the count.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true if num ` `// contains at least one digit ` `// that divides k ` `bool` `digitDividesK(``int` `num, ``int` `k) ` `{ ` `    ``while` `(num) { ` ` `  `        ``// Get the last digit ` `        ``int` `d = num % 10; ` ` `  `        ``// If the digit is non-zero ` `        ``// and it divides k ` `        ``if` `(d != 0 and k % d == 0) ` `            ``return` `true``; ` ` `  `        ``// Remove the last digit ` `        ``num = num / 10; ` `    ``} ` ` `  `    ``// There is no digit in num ` `    ``// that divides k ` `    ``return` `false``; ` `} ` ` `  `// Function to return the required ` `// count of elements from the given ` `// range which contain at least one ` `// digit that divides k ` `int` `findCount(``int` `l, ``int` `r, ``int` `k) ` `{ ` ` `  `    ``// To store the result ` `    ``int` `count = 0; ` ` `  `    ``// For every number from the range ` `    ``for` `(``int` `i = l; i <= r; i++) { ` ` `  `        ``// If any digit of the current ` `        ``// number divides k ` `        ``if` `(digitDividesK(i, k)) ` `            ``count++; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `l = 20, r = 35; ` `    ``int` `k = 45; ` ` `  `    ``cout << findCount(l, r, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` ` `  `class` `GFG ` `{ ` `    ``// Function that returns true if num  ` `    ``// contains at least one digit  ` `    ``// that divides k  ` `    ``static` `boolean` `digitDividesK(``int` `num, ``int` `k)  ` `    ``{  ` `        ``while` `(num != ``0``)  ` `        ``{  ` `     `  `            ``// Get the last digit  ` `            ``int` `d = num % ``10``;  ` `     `  `            ``// If the digit is non-zero  ` `            ``// and it divides k  ` `            ``if` `(d != ``0` `&& k % d == ``0``)  ` `                ``return` `true``;  ` `     `  `            ``// Remove the last digit  ` `            ``num = num / ``10``;  ` `        ``}  ` `     `  `        ``// There is no digit in num  ` `        ``// that divides k  ` `        ``return` `false``;  ` `    ``}  ` `     `  `    ``// Function to return the required  ` `    ``// count of elements from the given  ` `    ``// range which contain at least one  ` `    ``// digit that divides k  ` `    ``static` `int` `findCount(``int` `l, ``int` `r, ``int` `k)  ` `    ``{  ` `     `  `        ``// To store the result  ` `        ``int` `count = ``0``;  ` `     `  `        ``// For every number from the range  ` `        ``for` `(``int` `i = l; i <= r; i++)  ` `        ``{  ` `     `  `            ``// If any digit of the current  ` `            ``// number divides k  ` `            ``if` `(digitDividesK(i, k))  ` `                ``count++;  ` `        ``}  ` `        ``return` `count;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String []args) ` `    ``{  ` `        ``int` `l = ``20``, r = ``35``;  ` `        ``int` `k = ``45``;  ` `     `  `        ``System.out.println(findCount(l, r, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function that returns true if num ` `# contains at least one digit ` `# that divides k ` `def` `digitDividesK(num, k): ` `    ``while` `(num): ` ` `  `        ``# Get the last digit ` `        ``d ``=` `num ``%` `10` ` `  `        ``# If the digit is non-zero ` `        ``# and it divides k ` `        ``if` `(d !``=` `0` `and` `k ``%` `d ``=``=` `0``): ` `            ``return` `True` ` `  `        ``# Remove the last digit ` `        ``num ``=` `num ``/``/` `10` ` `  `    ``# There is no digit in num ` `    ``# that divides k ` `    ``return` `False` ` `  `# Function to return the required ` `# count of elements from the given ` `# range which contain at least one ` `# digit that divides k ` `def` `findCount(l, r, k): ` ` `  `    ``# To store the result ` `    ``count ``=` `0` ` `  `    ``# For every number from the range ` `    ``for` `i ``in` `range``(l, r ``+` `1``): ` ` `  `        ``# If any digit of the current ` `        ``# number divides k ` `        ``if` `(digitDividesK(i, k)): ` `            ``count ``+``=` `1` ` `  `    ``return` `count ` ` `  `# Driver code ` `l ``=` `20` `r ``=` `35` `k ``=` `45` ` `  `print``(findCount(l, r, k)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Function that returns true if num  ` `    ``// contains at least one digit  ` `    ``// that divides k  ` `    ``static` `bool` `digitDividesK(``int` `num, ``int` `k)  ` `    ``{  ` `        ``while` `(num != 0)  ` `        ``{  ` `     `  `            ``// Get the last digit  ` `            ``int` `d = num % 10;  ` `     `  `            ``// If the digit is non-zero  ` `            ``// and it divides k  ` `            ``if` `(d != 0 && k % d == 0)  ` `                ``return` `true``;  ` `     `  `            ``// Remove the last digit  ` `            ``num = num / 10;  ` `        ``}  ` `     `  `        ``// There is no digit in num  ` `        ``// that divides k  ` `        ``return` `false``;  ` `    ``}  ` `     `  `    ``// Function to return the required  ` `    ``// count of elements from the given  ` `    ``// range which contain at least one  ` `    ``// digit that divides k  ` `    ``static` `int` `findCount(``int` `l, ``int` `r, ``int` `k)  ` `    ``{  ` `     `  `        ``// To store the result  ` `        ``int` `count = 0;  ` `     `  `        ``// For every number from the range  ` `        ``for` `(``int` `i = l; i <= r; i++)  ` `        ``{  ` `     `  `            ``// If any digit of the current  ` `            ``// number divides k  ` `            ``if` `(digitDividesK(i, k))  ` `                ``count++;  ` `        ``}  ` `        ``return` `count;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main() ` `    ``{  ` `        ``int` `l = 20, r = 35;  ` `        ``int` `k = 45;  ` `     `  `        ``Console.WriteLine(findCount(l, r, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```10
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.