# Count n digit numbers divisible by given number

Given number of digit n and a number, the task is to count all the numbers which are divisible by that number and having n digit.

Examples :

```Input : n = 2, number = 7
Output : 9
There are nine n digit numbers that
are divisible by 7. Numbers are 14,
21, 28, 35, 42, 49, .... 70.

Input : n = 3, number = 7
Output : 128

Input : n = 4, number = 4
Output : 2250
```

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

Native Approach: Traverse through all n digit numbers. For every number check for divisibility,

## C++

 `// Simple CPP program to count n digit  ` `// divisible numbers. ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Returns count of n digit numbers ` `// divisible by 'number' ` `int` `numberofterm(``int` `n, ``int` `number) ` `{ ` `    ``// compute the first and last term ` `    ``int` `firstnum = ``pow``(10, n - 1); ` `    ``int` `lastnum = ``pow``(10, n); ` ` `  `    ``// count total number of which having ` `    ``// n digit and divisible by number ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = firstnum; i < lastnum; i++)  ` `        ``if` `(i % number == 0) ` `            ``count++;  ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 3, num = 7; ` `    ``cout << numberofterm(n, num) << ``"\n"``; ` `    ``return` `0; ` `} `

## Java

 `// Simple Java program to count n digit  ` `// divisible numbers. ` `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `    ``// Returns count of n digit numbers ` `    ``// divisible by 'number' ` `    ``static` `int` `numberofterm(``int` `n, ``int` `number) ` `    ``{ ` `        ``// compute the first and last term ` `        ``int` `firstnum = (``int``)Math.pow(``10``, n - ``1``); ` `        ``int` `lastnum = (``int``)Math.pow(``10``, n); ` `     `  `        ``// count total number of which having ` `        ``// n digit and divisible by number ` `        ``int` `count = ``0``; ` `        ``for` `(``int` `i = firstnum; i < lastnum; i++)  ` `            ``if` `(i % number == ``0``) ` `                ``count++;  ` `        ``return` `count; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `n = ``3``, num = ``7``; ` `        ``System.out.println(numberofterm(n, num)); ` `    ``} ` `} ` ` `  `// This code is contributed by Ajit. `

## Python3

 `# Simple Python 3 program to count n digit  ` `# divisible numbers ` ` `  `import` `math ` ` `  `# Returns count of n digit  ` `# numbers divisible by number ` `def` `numberofterm(n, number): ` ` `  `    ``# compute the first and last term ` `    ``firstnum ``=` `math.``pow``(``10``, n ``-` `1``) ` `    ``lastnum ``=` `math.``pow``(``10``, n) ` ` `  `    ``# count total number of which having ` `    ``# n digit and divisible by number ` `    ``count ``=` `0` `    ``for` `i ``in` `range``(``int``(firstnum), ``int``(lastnum)):  ` `        ``if` `(i ``%` `number ``=``=` `0``): ` `            ``count ``+``=` `1` `    ``return` `count ` ` `  ` `  `# Driver code ` `n ``=` `3`  `num ``=` `7` `print``(numberofterm(n, num)) ` ` `  `# This article is contributed ` `# by Smitha Dinesh Semwal `

## C#

 `// Simple C# program to count n digit  ` `// divisible numbers. ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Returns count of n digit numbers ` `    ``// divisible by 'number' ` `    ``static` `int` `numberofterm(``int` `n, ``int` `number) ` `    ``{ ` `        ``// compute the first and last term ` `        ``int` `firstnum = (``int``)Math.Pow(10, n - 1); ` `        ``int` `lastnum = (``int``)Math.Pow(10, n); ` `     `  `        ``// count total number of which having ` `        ``// n digit and divisible by number ` `        ``int` `count = 0; ` `        ``for` `(``int` `i = firstnum; i < lastnum; i++)  ` `            ``if` `(i % number == 0) ` `                ``count++;  ` `        ``return` `count; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `n = 3, num = 7; ` `        ``Console.Write(numberofterm(n, num)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal `

## PHP

 ` `

Output:

```128
```

Efficient Approach : Find the first and last terms divisible, then apply the below formula

Count of divisible = (lastnumber – firstnumber)/number + 1

## C++

 `// Efficient CPP program to count n digit  ` `// divisible numbers. ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// find the number of term ` `int` `numberofterm(``int` `digit, ``int` `number) ` `{ ` `    ``// compute the first and last term ` `    ``int` `firstnum = ``pow``(10, digit - 1); ` `    ``int` `lastnum = ``pow``(10, digit); ` ` `  `    ``// first number which is divisible by given number ` `    ``firstnum = (firstnum - firstnum % number) + number; ` ` `  `    ``// last number which is divisible by given number ` `    ``lastnum = (lastnum - lastnum % number); ` ` `  `    ``// Apply the formula here ` `    ``return` `((lastnum - firstnum) / number + 1); ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``int` `number = 7; ` `    ``cout << numberofterm(n, number) << ``"\n"``; ` `    ``return` `0; ` `} `

## Java

 `// Efficient Java program to count n digit  ` `// divisible numbers. ` `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `    ``// find the number of term ` `    ``static` `int` `numberofterm(``int` `digit, ``int` `number) ` `    ``{ ` `        ``// compute the first and last term ` `        ``int` `firstnum = (``int``)Math.pow(``10``, digit - ``1``); ` `        ``int` `lastnum = (``int``)Math.pow(``10``, digit); ` `     `  `        ``// first number which is divisible by given number ` `        ``firstnum = (firstnum - firstnum % number) + number; ` `     `  `        ``// last number which is divisible by given number ` `        ``lastnum = (lastnum - lastnum % number); ` `     `  `        ``// Apply the formula here ` `        ``return` `((lastnum - firstnum) / number + ``1``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `n = ``3``; ` `        ``int` `number = ``7``; ` `        ``System.out.println(numberofterm(n, number)); ` `    ``} ` `} ` ` `  `// This code is contributed by Ajit. `

## Python3

 `# Efficient Python program to   ` `# count n digit divisible numbers. ` ` `  `# Find the number of term ` `def` `numberofterm(digit, number): ` `     `  `    ``# compute the first and last term ` `    ``firstnum ``=` `pow``(``10``, digit ``-` `1``) ` `    ``lastnum ``=` `pow``(``10``, digit) ` ` `  `    ``# First number which is divisible by given number ` `    ``firstnum ``=` `(firstnum ``-` `firstnum ``%` `number) ``+` `number ` ` `  `    ``# last number which is divisible by given number ` `    ``lastnum ``=` `(lastnum ``-` `lastnum ``%` `number) ` ` `  `    ``# Apply the formula here ` `    ``return` `((lastnum ``-` `firstnum) ``/``/` `number ``+` `1``); ` ` `  `# Driver code ` `n ``=` `3``; number ``=` `7` `print``(numberofterm(n, number)) ` ` `  `# This code is contributed by Ajit. `

## C#

 `// Efficient C# program to count n digit  ` `// divisible numbers. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// find the number of term ` `    ``static` `int` `numberofterm(``int` `digit,  ` `                            ``int` `number) ` `    ``{ ` `         `  `        ``// compute the first and  ` `        ``// last term ` `        ``int` `firstnum = (``int``)Math.Pow(10, ` `                             ``digit - 1); ` `                              `  `        ``int` `lastnum = (``int``)Math.Pow(10,  ` `                                ``digit); ` `     `  `        ``// first number which is divisible ` `        ``// by given number ` `        ``firstnum = (firstnum - firstnum  ` `                      ``% number) + number; ` `     `  `        ``// last number which is divisible ` `        ``// by given number ` `        ``lastnum = (lastnum - lastnum ` `                              ``% number); ` `     `  `        ``// Apply the formula here ` `        ``return` `((lastnum - firstnum) ` `                          ``/ number + 1); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `n = 3; ` `        ``int` `number = 7; ` `         `  `        ``Console.WriteLine( ` `             ``numberofterm(n, number)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## PHP

 ` `

Output:

```128
```

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