# Times required by Simple interest for the Principal to become Y times itself

Given that a certain amount of money becomes **T1** times itself in **N1** years. The task is to find the number of years i.e. **N2** so that the amount becomes **T2** times itself at the same rate of simple interest.

**Examples:**

Input:T1 = 5, N1 = 7, T2 = 25Output:42

Input:T1 = 3, N1 = 5, T2 = 6Output:12.5

**Approach:**

Let us consider the 1st Example where T1 = 5, N1 = 7, T2 = 25

Now, Let P principal becomes 5P i.e (T1 * P) then Simple interest received is 4P.

(As S.I = Amount – P)

Now, in the second case, P has become 25P i.e (T2 * P) then simple interest received is 24P.

Now if we received 4P interest in N1 i.e 7 years then we will get an interest of 24P

in 7 * 6 years i.e in 42 years.

**Formula:**

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the no. of years` `float` `noOfYears(` `int` `t1, ` `int` `n1, ` `int` `t2)` `{` ` ` `float` `years = ((t2 - 1) * n1 / (` `float` `)(t1 - 1));` ` ` `return` `years;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `T1 = 3, N1 = 5, T2 = 6;` ` ` `cout << noOfYears(T1, N1, T2);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `class` `GFG` `{` ` ` `// Function to return the no. of years` `static` `float` `noOfYears(` `int` `t1, ` `int` `n1, ` `int` `t2)` `{` ` ` `float` `years = ((t2 - ` `1` `) * n1 / (` `float` `)(t1 - ` `1` `));` ` ` `return` `years;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `T1 = ` `3` `, N1 = ` `5` `, T2 = ` `6` `;` ` ` `System.out.println(noOfYears(T1, N1, T2));` `}` `}` `// This code is contributed by Code_Mech` |

## Python3

`# Python3 implementation of the approach` `# Function to return the no. of years` `def` `noOfYears(t1, n1, t2):` ` ` `years ` `=` `(t2 ` `-` `1` `) ` `*` `n1 ` `/` `(t1 ` `-` `1` `)` ` ` `return` `years` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `T1, N1, T2 ` `=` `3` `, ` `5` `, ` `6` ` ` `print` `(noOfYears(T1, N1, T2))` ` ` `# This code is contributed` `# by Rituraj Jain` |

## C#

`// C# implementation for above approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to return the no. of years` `static` `float` `noOfYears(` `int` `t1, ` `int` `n1, ` `int` `t2)` `{` ` ` `float` `years = ((t2 - 1) * n1 / (` `float` `)(t1 - 1));` ` ` `return` `years;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `T1 = 3, N1 = 5, T2 = 6;` ` ` `Console.WriteLine(noOfYears(T1, N1, T2));` `}` `}` `/* This code contributed by PrinciRaj1992 */` |

## PHP

`<?php` `// PHP implementation for above approach` ` ` `// Function to return the no. of years` `function` `noOfYears(` `$t1` `, ` `$n1` `, ` `$t2` `)` `{` ` ` `$years` `= ((` `$t2` `- 1) * ` `$n1` `/ (` `$t1` `- 1));` ` ` `return` `$years` `;` `}` `// Driver code` `$T1` `= 3;` `$N1` `= 5;` `$T2` `= 6;` `print` `(noOfYears(` `$T1` `, ` `$N1` `, ` `$T2` `));` `// This code contributed by mits` `?>` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to return the no. of years` `function` `noOfYears(t1, n1, t2)` `{` ` ` `var` `years = ((t2 - 1) * n1 / (t1 - 1));` ` ` `return` `years;` `}` `// Driver code` `var` `T1 = 3, N1 = 5, T2 = 6;` `document.write( noOfYears(T1, N1, T2));` `// This code is contributed by rutvik_56.` `</script>` |

**Output:**

12.5

**Time Complexity: **O(1)

**Auxiliary Space:** O(1)