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Third last digit in 5^N for given N
• Last Updated : 01 Apr, 2019

Given a positive integer N. The task is to find the value of 3rd digit from last (right-most) of 5N.

Examples:

```Input : N = 6
Output : 6
Explanation : Value of 56 = 15625.

Input : N = 3
Output : 1
Explanation : Value of 53 = 125.
```

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

Approach : Before moving to actual approach, some facts regarding number theory are listed below as:

• 53 is the smallest 3 digit number which is power of 5.
• As 125 * 5 = 625, this conclude that multiple of number (ending with 125) with 5 always construct 625 as last three digit of result.
• Again as, 625 * 5 = 3125, this conclude that multiple of number (ending with 625) with 5 always construct 125 as last three digit of result.

Hence, the final general solution is :

case 1: if n < 3, answer = 0.
case 2: if n >= 3 and is even , answer = 6.
case 3: if n >= 3 and is odd , answer = 1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach`` ` `#include ``using` `namespace` `std;`` ` `// Function to find the element``int` `findThirdDigit(``int` `n)``{``    ``// if n < 3``    ``if` `(n < 3)``        ``return` `0;`` ` `    ``// If n is even return 6``    ``// If n is odd return 1``    ``return` `n & 1 ? 1 : 6;``}`` ` `// Driver code``int` `main()``{``    ``int` `n = 7;`` ` `    ``cout << findThirdDigit(n);``  ` `    ``return` `0;``} `

## Java

 `// Java implementation of the ``// above approach``class` `GFG``{``     ` `// Function to find the element``static` `int` `findThirdDigit(``int` `n)``{``    ``// if n < 3``    ``if` `(n < ``3``)``        ``return` `0``;`` ` `    ``// If n is even return 6``    ``// If n is odd return 1``    ``return` `(n & ``1``) > ``0` `? ``1` `: ``6``;``}`` ` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `n = ``7``;`` ` `    ``System.out.println(findThirdDigit(n));``} ``}`` ` `// This code is contributed ``// by Akanksha Rai`

## Python3

 `# Python3 implementation of the ``# above approach`` ` `# Function to find the element``def` `findThirdDigit(n):`` ` `    ``# if n < 3``    ``if` `n < ``3``:``        ``return` `0`` ` `    ``# If n is even return 6``    ``# If n is odd return 1``    ``return` `1` `if` `n ``and` `1` `else` `6`` ` `# Driver code``n ``=` `7``print``(findThirdDigit(n))`` ` `# This code is contributed ``# by Shrikant13`

## C#

 `// C# implementation of the above approach``using` `System;`` ` `class` `GFG``{``     ` `// Function to find the element``static` `int` `findThirdDigit(``int` `n)``{``    ``// if n < 3``    ``if` `(n < 3)``        ``return` `0;`` ` `    ``// If n is even return 6``    ``// If n is odd return 1``    ``return` `(n & 1)>0 ? 1 : 6;``}`` ` `// Driver code``static` `void` `Main()``{``    ``int` `n = 7;`` ` `    ``Console.WriteLine(findThirdDigit(n));``} ``}`` ` `// This code is contributed by mits`

## PHP

 ``
Output:
```1
```

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