# Counting numbers whose difference from reverse is a product of k

Given two number l and r. Count the total numbers between l and r which when subtracted from their respective reverse, the difference is a product of k.

Examples:

```Input : 20 23 6
Output : 2
20 and 22 are the two numbers.
|20-2| = 18 which is a product of 6
|22-22| = 0 which is a product of 6

Input : 35 45 5
Output : 2
38 and 44 are the two numbers.
|38-83| = 45 which is a product of 5
|44-44| = 0 which is a product of 5
```

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

Approach: For each number between the given range check if the difference between the number and its reverse number is divisible by k, increment count if yes.

## C++

 `// C++ program to Count the numbers ` `// within a given range in which when ` `// you subtract a number from its ` `// reverse, the difference is a product ` `// of k ` `#include ` `using` `namespace` `std; ` ` `  `bool` `isRevDiffDivisible(``int` `x, ``int` `k) ` `{  ` `    ``// function to check if the number ` `    ``// and its reverse have their  ` `    ``// absolute difference divisible by k ` `    ``int` `n = x; ` `    ``int` `m = 0; ` `    ``int` `flag; ` `    ``while` `(x > 0)  ` `    ``{    ` `        ``// reverse the number ` `        ``m = m * 10 + x % 10; ` `        ``x /= 10; ` `    ``} ` `     `  `    ``return` `(``abs``(n - m) % k == 0); ` `} ` ` `  `int` `countNumbers(``int` `l, ``int` `r, ``int` `k) ` `{ ` `    ``int` `count = 0;  ` `    ``for` `(``int` `i = l; i <= r; i++)     ` `        ``if` `(isRevDiffDivisible(i, k))         ` `            ``++count;  ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `l = 20, r = 23, k = 6; ` `    ``cout << countNumbers(l, r, k) << endl;  ` `    ``return` `0; ` `} `

## Java

 `// Java program to Count the  ` `// numbers within a given range  ` `// in which when you subtract  ` `// a number from its reverse,  ` `// the difference is a product of k ` `import` `java.io.*; ` `import` `java.math.*; ` ` `  `class` `GFG { ` `     `  `    ``static` `boolean` `isRevDiffDivisible(``int` `x, ``int` `k) ` `    ``{  ` `        ``// function to check if the number ` `        ``// and its reverse have their  ` `        ``// absolute difference divisible by k ` `        ``int` `n = x; ` `        ``int` `m = ``0``; ` `        ``int` `flag; ` `        ``while` `(x > ``0``)  ` `        ``{  ` `            ``// reverse the number ` `            ``m = m * ``10` `+ x % ``10``; ` `            ``x /= ``10``; ` `        ``} ` `         `  `        ``return` `(Math.abs(n - m) % k == ``0``); ` `    ``} ` `     `  `    ``static` `int` `countNumbers(``int` `l, ``int` `r, ``int` `k) ` `    ``{ ` `        ``int` `count = ``0``;  ` `        ``for` `(``int` `i = l; i <= r; i++)  ` `            ``if` `(isRevDiffDivisible(i, k))      ` `                ``++count;  ` `        ``return` `count; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `l = ``35``, r = ``45``, k = ``5``; ` `        ``System.out.println(countNumbers(l, r, k)); ` `    ``} ` `} ` ` `  `// This code is contributed  ` `// by Nikita Tiwari. `

## Python3

 `# Python 3 program to Count the numbers ` `# within a given range in which when you ` `# subtract a number from its reverse, ` `# the difference is a product of k ` ` `  `def` `isRevDiffDivisible(x, k) : ` `    ``# function to check if the number ` `    ``# and its reverse have their  ` `    ``# absolute difference divisible by k ` `    ``n ``=` `x; m ``=` `0` `    ``while` `(x > ``0``) : ` ` `  `        ``# Reverse the number ` `        ``m ``=` `m ``*` `10` `+` `x ``%` `10` `        ``x ``=` `x ``/``/` `10` `         `  `    ``return` `(``abs``(n ``-` `m) ``%` `k ``=``=` `0``) ` ` `  `def` `countNumbers(l, r, k) : ` `    ``count ``=` `0` `    ``for` `i ``in` `range``(l, r ``+` `1``) : ` ` `  `        ``if` `(isRevDiffDivisible(i, k)) : ` `            ``count ``=` `count ``+` `1` `     `  `    ``return` `count ` `     `  `     `  `# Driver code ` `l ``=` `20``; r ``=` `23``; k ``=` `6` `print``(countNumbers(l, r, k)) ` ` `  ` `  `# This code is contributed by Nikita Tiwari. `

## C#

 `// C# program to Count the  ` `// numbers within a given range  ` `// in which when you subtract  ` `// a number from its reverse,  ` `// the difference is a product of k ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `bool` `isRevDiffDivisible(``int` `x, ``int` `k) ` `    ``{  ` `        ``// function to check if the number ` `        ``// and its reverse have their  ` `        ``// absolute difference divisible by k ` `        ``int` `n = x; ` `        ``int` `m = 0; ` `         `  `        ``// int flag; ` `        ``while` `(x > 0)  ` `        ``{  ` `            ``// reverse the number ` `            ``m = m * 10 + x % 10; ` `            ``x /= 10; ` `        ``} ` `         `  `        ``return` `(Math.Abs(n - m) % k == 0); ` `    ``} ` `     `  `    ``static` `int` `countNumbers(``int` `l, ``int` `r, ``int` `k) ` `    ``{ ` `        ``int` `count = 0;  ` `         `  `        ``for` `(``int` `i = l; i <= r; i++)  ` `            ``if` `(isRevDiffDivisible(i, k))  ` `                ``++count;  ` `        ``return` `count; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `l = 35, r = 45, k = 5; ` `        ``Console.WriteLine(countNumbers(l, r, k)); ` `    ``} ` `} ` ` `  `// This code is contributed  ` `// by vt_m. `

## PHP

 ` 0)  ` `    ``{  ` `        ``// reverse the number ` `        ``\$m` `= ``\$m` `* 10 + ``\$x` `% 10; ` `        ``\$x` `= (int)``\$x` `/ 10; ` `    ``} ` `     `  `    ``return` `(``abs``(``\$n` `- ``\$m``) % ` `                ``\$k` `== 0); ` `} ` ` `  `function` `countNumbers(``\$l``, ``\$r``, ``\$k``) ` `{ ` `    ``\$count` `= 0;  ` `    ``for` `(``\$i` `= ``\$l``; ``\$i` `<= ``\$r``; ``\$i``++)  ` `        ``if` `(isRevDiffDivisible(``\$i``, ``\$k``))  ` `            ``++``\$count``;  ` `    ``return` `\$count``; ` `} ` ` `  `// Driver code ` `\$l` `= 20; ``\$r` `= 23; ``\$k` `= 6; ` `echo` `countNumbers(``\$l``, ``\$r``, ``\$k``);  ` ` `  `// This code is contributed by mits. ` `?> `

Output:

```2
```

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 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.

Improved By : Mithun Kumar

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.