# Program to find the number of men initially

A certain number of men can do a certain piece of work in **D** days. If there were **m** more men engaged in the work then the work can be done in **d** days less. The task is to find how many men were there initially.**Examples:**

Input:D = 5, m = 4, d = 4Output:1Input:D = 180, m = 30, d = 20Output:240

**Approach:** Let the initial number of men be **M** and days be **D**

Amount of work completed **M** men in **D** days will be **M * D**

i.e. **Work Done = M * D** …(1)

If there are **M + m** men then the same amount of work is completed in **D – d** days.

i.e. **Work Done = (M + m) * (D – d)** …(2)

Equating equations 1 and 2,

M * D = (M + m) * (D – d)

M * D = M * (D – d) + m * (D – d)

M * D – M * (D – d) = m * (D – d)

M * (D – (D – d)) = m * (D – d)M = m * (D – d) / d

Below is the implementation of the above approach:

## C++

`// C++ implementation of above approach.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the` `// number of men initially` `int` `numberOfMen(` `int` `D, ` `int` `m, ` `int` `d)` `{` ` ` `int` `Men = (m * (D - d)) / d;` ` ` `return` `Men;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `D = 5, m = 4, d = 4;` ` ` `cout << numberOfMen(D, m, d);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach ` `import` `java.util.*;` `class` `GFG ` `{ ` ` ` `// Function to return the ` `// number of men initially ` `static` `int` `numberOfMen(` `int` `D, ` `int` `m, ` `int` `d) ` `{ ` ` ` `int` `Men = (m * (D - d)) / d; ` ` ` `return` `Men; ` `} ` `// Driver code ` `public` `static` `void` `main(String args[])` `{ ` ` ` `int` `D = ` `5` `, m = ` `4` `, d = ` `4` `; ` ` ` `System.out.println(numberOfMen(D, m, d)); ` `} ` `} ` `// This code is contributed by Arnab Kundu` |

## Python3

`# Python implementation of above approach. ` `# Function to return the ` `# number of men initially ` `def` `numberOfMen(D, m, d):` ` ` `Men ` `=` `(m ` `*` `(D ` `-` `d)) ` `/` `d; ` ` ` `return` `int` `(Men); ` `# Driver code ` `D ` `=` `5` `; m ` `=` `4` `; d ` `=` `4` `; ` `print` `(numberOfMen(D, m, d)); ` `# This code contributed by Rajput-Ji` |

## C#

`// C# implementation of the approach ` `using` `System;` `class` `GFG ` `{ ` ` ` `// Function to return the ` `// number of men initially ` `static` `int` `numberOfMen(` `int` `D, ` `int` `m, ` `int` `d) ` `{ ` ` ` `int` `Men = (m * (D - d)) / d; ` ` ` `return` `Men; ` `} ` `// Driver code ` `public` `static` `void` `Main()` `{ ` ` ` `int` `D = 5, m = 4, d = 4; ` ` ` `Console.WriteLine(numberOfMen(D, m, d)); ` `} ` `} ` `// This code is contributed by anuj_67..` |

## Javascript

`<script>` `// Javascript implementation of above approach.` `// Function to return the` `// number of men initially` `function` `numberOfMen(D, m, d)` `{` ` ` `var` `Men = (m * (D - d)) / d;` ` ` `return` `Men;` `}` `// Driver code` `var` `D = 5, m = 4, d = 4;` `document.write(numberOfMen(D, m, d));` `</script>` |

**Output:**

1