# Count of matrices (of different orders) with given number of elements

Given a number N denotes the total number of elements in a matrix, the task is to print all possible order of matrix. An order is a pair (m, n) of integers where m is number of rows and n is number of columns. For example, if the number of elements is 8 then all possible orders are:
(1, 8), (2, 4), (4, 2), (8, 1).

Examples:

Input: N = 8
Output: (1, 2) (2, 4) (4, 2) (8, 1)

Input: N = 100
Output:
(1, 100) (2, 50) (4, 25) (5, 20) (10, 10) (20, 5) (25, 4) (50, 2) (100, 1)

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

Approach:
A matrix is said to be of order m x n if it has m rows and n columns. The total number of elements in a matrix is equal to (m*n). So we start from 1 and check one by one if it divides N(the total number of elements). If it divides, it will be one possible order.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print all possible order ` `void` `printAllOrder(``int` `n) ` `{ ` `    ``// total number of elements in a matrix  ` `    ``// of order m * n is equal (m*n) ` `    ``// where m is number of rows and n is  ` `    ``// number of columns ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` ` `  `        ``// if n is divisible by i then i ` `        ``// and n/i will be the one ` `        ``// possible order of the matrix ` `        ``if` `(n % i == 0) { ` ` `  `            ``// print the given format ` `            ``cout << i << ``" "` `<< n / i << endl; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 10; ` `    ``printAllOrder(n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` ` `  ` `  `class` `GFG ` `    ``{ ` `    ``// Function to print all possible order ` `    ``static` `void` `printAllOrder(``int` `n) ` `    ``{ ` `        ``// total number of elements in a matrix  ` `        ``// of order m * n is equal (m*n) ` `        ``// where m is number of rows and n is  ` `        ``// number of columns ` `        ``for` `(``int` `i = ``1``; i <= n; i++) { ` `     `  `            ``// if n is divisible by i then i ` `            ``// and n/i will be the one ` `            ``// possible order of the matrix ` `            ``if` `(n % i == ``0``) { ` `     `  `                ``// print the given format ` `                ``System.out.println( i + ``" "` `+ n / i ); ` `            ``} ` `        ``} ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `        ``int` `n = ``10``; ` `        ``printAllOrder(n); ` `         `  `    ``} ` ` `  `} ` ` `  ` `  `// This code is contributed by ihritik `

## Python

 `# Python implementation of the above approach ` ` `  `# Function to print all possible order ` `def` `printAllOrder(n): ` ` `  `    ``# total number of elements in a matrix  ` `    ``# of order m * n is equal (m*n) ` `    ``# where m is number of rows and n is  ` `    ``# number of columns ` `    ``for` `i ``in` `range``(``1``,n``+``1``): ` ` `  `        ``# if n is divisible by i then i ` `        ``# and n/i will be the one ` `        ``# possible order of the matrix ` `        ``if` `(n ``%` `i ``=``=` `0``) : ` ` `  `            ``# print the given format ` `            ``print``( i ,n ``/``/` `i ) ` `         `  `     `  ` `  ` `  `# Driver code ` `n ``=` `10` `printAllOrder(n) ` ` `  ` `  `# This code is contributed by ihritik `

## C#

 `// C# implementation of the above approach ` ` `  `using` `System; ` `class` `GFG ` `    ``{ ` `    ``// Function to print all possible order ` `    ``static` `void` `printAllOrder(``int` `n) ` `    ``{ ` `        ``// total number of elements in a matrix  ` `        ``// of order m * n is equal (m*n) ` `        ``// where m is number of rows and n is  ` `        ``// number of columns ` `        ``for` `(``int` `i = 1; i <= n; i++) { ` `     `  `            ``// if n is divisible by i then i ` `            ``// and n/i will be the one ` `            ``// possible order of the matrix ` `            ``if` `(n % i == 0) { ` `     `  `                ``// print the given format ` `                ``Console.WriteLine( i + ``" "` `+ n / i ); ` `            ``} ` `        ``} ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 10; ` `        ``printAllOrder(n); ` `         `  `    ``} ` ` `  `} ` ` `  `// This code is contributed by ihritik `

## PHP

 ` `

Output:

```1 10
2 5
5 2
10 1
```

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