Third last digit in 5^N for given N

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

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

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

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

• 5³ is the smallest 3-digit number which is a power of 5.
• As 125 * 5 = 625, this concludes that multiple of numbers (ending with 125) with 5 always construct 625 as of the last three digits of the result.
• Again as, 625 * 5 = 3125, this concludes that multiple numbers (ending with 625) with 5 always construct 125 as the last three digits of the 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 <bits/stdc++.h>
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
import java.io.*;
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

<?php
// PHP implementation of the above approach

// Function to find the element
function findThirdDigit($n)
{
    // if n < 3
    if ($n < 3)
        return 0;

    // If n is even return 6
    // If n is odd return 1
    return $n & 1 ? 1 : 6;
}

// Driver code
$n = 7;
echo findThirdDigit($n);

// This code contributed by
// PrinciRaj1992
?>

Javascript

<script>

 // Javascript implementation of the above approach

// Function to find the element
function findThirdDigit(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
var n = 7;
document.write( findThirdDigit(n));

</script>

Output:

1

Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.