Last two digits of powers of 7

Given a positive N, the task is to find the last two digits of 7^N.
Examples: 
 

Input: N = 5 
Output: 07 
Explanation: 
The value of 7⁵ = 7 * 7 * 7 * 7 * 7 = 8507 
Therefore, the last two digits are 07.
Input: N = 12 
Output: 01 
Explanation: 
The value of 7¹² = 13841287201 
Therefore, the last two digits are 01. 
 

Approach: A general approach to finding the last K digits of X^Y is to discuss this article in logarithmic time complexity. In this article, we will discuss the constant time solution.
Below is the observation for the value of 7^N for some values of N: 
 

7¹ = 7 last two digit = 07 
7² = 49 last two digit = 49 
7³ = 243 last two digit = 43 
7⁴ = 2401 last two digit = 01
7⁵ = 16807 last two digit = 07 
7⁶ = 117649 last two digit = 49 
7⁷ = 823543 last two digit = 43 
7⁸ = 5764801 last two digit = 01 
 

Based on the above observations we have the following cases: 
 

1. If the last two digit in 7^N = 07 when N = 4K + 3.
2. If the last two digit in 7^N = 49 when N = 4K + 2.
3. If the last two digit in 7^N = 43 when N = 4K + 1.
4. If the last two digit in 7^N = 01 when N = 4K.

Below is the implementation of the above approach:
 

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the last
// two digits of 7^N
string get_last_two_digit(int N)
{
    // Case 4
    if (N % 4 == 0)
        return "01";

    // Case 3
    else if (N % 4 == 1)
        return "07";

    // Case 2
    else if (N % 4 == 2)
        return "49";

    // Case 1
    return "43";
}

// Driver Code
int main()
{
    // Given Number
    int N = 12;

    // Function Call
    cout << get_last_two_digit(N);
    return 0;
}

// Java program for the above approach 
import java.io.*;
import java.util.*;

class GFG{

// Function to find the last 
// two digits of 7^N 
public static String get_last_two_digit(int N)
{
    // Case 4
    if (N % 4 == 0)
        return "01";

    // Case 3
    else if (N % 4 == 1)
        return "07";

    // Case 2
    else if (N % 4 == 2)
        return "49";

    // Case 1
    return "43";
}

// Driver code
public static void main(String[] args)
{
    int N = 12;

    // Function Call
    System.out.println(get_last_two_digit(N));
}
}

// This code is contributed by grand_master

# Python3 program for the above approach

# Function to find the last
# two digits of 7 ^ N
def get_last_two_digit(N):

    # Case 4
    if (N % 4 == 0):
        return "01";

    # Case 3
    elif (N % 4 == 1):
        return "07";

    # Case 2
    elif (N % 4 == 2):
        return "49";

    # Case 1
    return "43";
    
# Driver Code

# Given number
N = 12;

# Function call
print( get_last_two_digit(N))

# This code is contributed by grand_master

// C# program for the above approach
using System;

namespace GFG{
class GFG{
    
// Function to find the last 
// two digits of 7^N 
public static String get_last_two_digit(int N)
{
    // Case 4
    if (N % 4 == 0)
        return "01";

    // Case 3
    else if (N % 4 == 1)
        return "07";

    // Case 2
    else if (N % 4 == 2)
        return "49";

    // Case 1
    return "43";
}

// Driver code
public static void Main()
{
    
    // Given number
    int N = 12;

    // Function Call
    Console.Write(get_last_two_digit(N));
}
}
}

// This code is contributed by grand_master

<script>

// Javascript program for the above approach 

// Function to find the last
    // two digits of 7^N
    function get_last_two_digit(N) 
    {
        // Case 4
        if (N % 4 == 0)
            return "01";

        // Case 3
        else if (N % 4 == 1)
            return "07";

        // Case 2
        else if (N % 4 == 2)
            return "49";

        // Case 1
        return "43";
    }

    // Driver code
    
        var N = 12;

        // Function Call
        document.write(get_last_two_digit(N));

// This code contributed by gauravrajput1 

</script>

01

Time Complexity: O(1) 
Auxiliary Space: O(1)