Given two numbers m and n, count number of m digit numbers that are divisible by n.

Examples:

Input : m = 2 n = 6 Output : 15 Two digit numbers that are divisible by 6 are 12, 18, 24, 30, 36, ....., 96. Input : m = 3 n = 5 Output :180

A **simple solution** is two try all m digit numbers. For every number, check if it is divisible by n. If yes, we increment count.

An **efficient solution** involves following steps.

The idea is based on the fact that starting from first divisible number, every n-th number is divisible by n.

- Find largest m digit number.
- Find largest m-1 digit number.
- Divide both number by n and subtract later from prior.

Below is the implementation of above steps.

## C++

`// C++ program to count m digit numbers having` `// n as divisor.` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Returns count of m digit numbers having n` `// as divisor` `int` `findCount(` `int` `m, ` `int` `n)` `{ ` ` ` `// generating largest number of m digit` ` ` `int` `num1 = 0;` ` ` `for` `(` `int` `i = 0; i < m; i++)` ` ` `num1 = (num1 * 10) + 9;` ` ` `// generating largest number of m-1 digit` ` ` `int` `num2 = 0;` ` ` `for` `(` `int` `i = 0; i < (m - 1); i++)` ` ` `num2 = (num2 * 10) + 9;` ` ` `// returning number of dividend` ` ` `return` `((num1 / n) - (num2 / n));` `}` `// Driver code` `int` `main()` `{` ` ` `int` `m = 2, n = 6;` ` ` `printf` `(` `"%d\n"` `, findCount(m, n));` ` ` `return` `0;` `}` |

## Java

`// Java program to count m digit numbers having` `// n as divisor.` `class` `Main` `{` ` ` `// Returns count of m digit numbers having n` ` ` `// as divisor` ` ` `static` `int` `findCount(` `int` `m, ` `int` `n)` ` ` `{ ` ` ` `// generating largest number of m digit` ` ` `int` `num1 = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < m; i++)` ` ` `num1 = (num1 * ` `10` `) + ` `9` `;` ` ` ` ` `// generating largest number of m-1 digit` ` ` `int` `num2 = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < (m - ` `1` `); i++)` ` ` `num2 = (num2 * ` `10` `) + ` `9` `;` ` ` ` ` `// returning number of dividend` ` ` `return` `((num1 / n) - (num2 / n));` ` ` `}` ` ` ` ` `// main function` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `m = ` `2` `, n = ` `6` `;` ` ` `System.out.println(findCount(m, n));` ` ` `}` `}` `/* This code is contributed by Harsh Agarwal */` |

## Python3

`# Python3 program to count m digit` `# numbers having n as divisor.` `# Returns count of m digit` `# numbers having n as divisor` `def` `findCount(m, n):` ` ` `# Generating largest number of m digit` ` ` `num1 ` `=` `0` ` ` ` ` `for` `i ` `in` `range` `(` `0` `, m):` ` ` `num1 ` `=` `(num1 ` `*` `10` `) ` `+` `9` ` ` `# Generating largest number of m-1 digit` ` ` `num2 ` `=` `0` ` ` ` ` `for` `i ` `in` `range` `(` `0` `, (m ` `-` `1` `)):` ` ` `num2 ` `=` `(num2 ` `*` `10` `) ` `+` `9` ` ` `# returning number of dividend` ` ` `return` `int` `((num1 ` `/` `n) ` `-` `(num2 ` `/` `n))` `# Driver code` `m ` `=` `2` `; n ` `=` `6` `print` `(findCount(m, n))` `# This code is contributed by Smitha Dinesh Semwal` |

## C#

`// C# program to count m digit numbers` `// having n as divisor.` `using` `System;` `class` `GfG {` ` ` ` ` `// Returns count of m digit numbers` ` ` `// having n as divisor` ` ` `static` `int` `findCount(` `int` `m, ` `int` `n)` ` ` `{` ` ` ` ` `// generating largest number` ` ` `// of m digit` ` ` `int` `num1 = 0;` ` ` `for` `(` `int` `i = 0; i < m; i++)` ` ` `num1 = (num1 * 10) + 9;` ` ` ` ` `// generating largest number` ` ` `// of m-1 digit` ` ` `int` `num2 = 0;` ` ` `for` `(` `int` `i = 0; i < (m - 1); i++)` ` ` `num2 = (num2 * 10) + 9;` ` ` ` ` `// returning number of dividend` ` ` `return` `((num1 / n) - (num2 / n));` ` ` `}` ` ` ` ` `// main function` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `m = 2, n = 6;` ` ` ` ` `Console.Write(findCount(m, n));` ` ` `}` `}` `// This code is contributed by parashar.` |

## PHP

`<?php` `// PHP program to count m digit` `// numbers having n as divisor.` `// Returns count of m digit numbers` `// having n as divisor` `function` `findCount(` `$m` `, ` `$n` `)` `{` ` ` `// generating largest number` ` ` `// of m digit` ` ` `$num1` `= 0;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$m` `; ` `$i` `++)` ` ` `$num1` `= (` `$num1` `* 10) + 9;` ` ` `// generating largest number` ` ` `// of m-1 digit` ` ` `$num2` `= 0;` ` ` `for` `(` `$i` `= 0; ` `$i` `< (` `$m` `- 1); ` `$i` `++)` ` ` `$num2` `= (` `$num2` `* 10) + 9;` ` ` `// returning number of dividend` ` ` `return` `((` `$num1` `/ ` `$n` `) - (` `$num2` `/ ` `$n` `));` `}` `// Driver code` `$m` `= 2; ` `$n` `= 6;` `echo` `findCount(` `$m` `, ` `$n` `), ` `"\n"` `;` `// This code is contributed by ajit` `?>` |

## Javascript

`<script>` `// Javascript program to count m digit` `// numbers having n as divisor.` `// Returns count of m digit numbers` `// having n as divisor` `function` `findCount(m, n)` `{` ` ` `// generating largest number` ` ` `// of m digit` ` ` `let num1 = 0;` ` ` `for` `(let i = 0; i < m; i++)` ` ` `num1 = (num1 * 10) + 9;` ` ` `// generating largest number` ` ` `// of m-1 digit` ` ` `let num2 = 0;` ` ` `for` `(let i = 0; i < (m - 1); i++)` ` ` `num2 = (num2 * 10) + 9;` ` ` `// returning number of dividend` ` ` `return` `((num1 / n) - (num2 / n));` `}` `// Driver code` `let m = 2; n = 6;` `document.write(findCount(m, n) + ` `"<br>"` `);` `// This code is contributed by gfgking` `</script>` |

Output :

15

Time complexity : O(m)

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