# Count of m digit integers that are divisible by an integer n

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
```

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

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.

1. Find largest m digit number.
2. Find largest m-1 digit number.
3. 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 ` `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

 ` `

Output :

```
15

```

Time complexity : O(m)

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Improved By : parashar, jit_t

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