Find value of (1^n + 2^n + 3^n + 4^n ) mod 5

You are given an positive integer n. You have to find the value of (1ⁿ +2ⁿ + 3ⁿ + 4ⁿ ) mod 5. 
Note : Value of n may be very large of order 10¹⁵. 
Examples: 

Input : n = 4
Output : 4
Explanation : (1⁴ + 2⁴ + 3⁴ + 4⁴)mod 5 = (1+16+81+256)mod 5 = 354 mod 5 = 4

Input : n = 2
Output : 0
Explanation : (1² + 2² + 3² + 4²)mod 5 = (1+4+9+16)mod 5 = 30 mod 5 = 0

Basic Approach : If you will solve this question with a very basic approach of finding value of (1ⁿ +2ⁿ + 3ⁿ + 4ⁿ ) and then finding its modulo value for 5, you will certainly get your answer but for the larger value of n we must got wrong answer as you will be unable to store value of (1ⁿ +2ⁿ + 3ⁿ + 4ⁿ ) properly. 

Better and Proper Approach : Before proceeding to solution lets go through some of periodical properties of power of 2, 3 & 4. 

• f(n) = 2ⁿ is periodical for n = 4 in terms of last digit. i.e. last digit of 2ⁿ always repeat for next 4th value of n. (ex: 2, 4, 8, 16, 32, 64…)
• f(n) = 3ⁿ is periodical for n = 4 in terms of last digit. i.e. last digit of 3ⁿ always repeat for next 4th value of n.(ex: 3, 9, 27, 81, 243, 729…)
• f(n) = 4ⁿ is periodical for n = 2 in terms of last digit. i.e. last digit of 4ⁿ always repeat for next 2nd value of n.(ex: 4, 16, 64, 256..)
• 1ⁿ is going to be 1 always, independent of n.

So, If we will have a close look for periodicity of f(n) = (1ⁿ +2ⁿ + 3ⁿ + 4ⁿ ) we will get that its periodicity is also 4 and its last digits occurs as : 

• for n = 1, f(n) = 10
• for n = 2, f(n) = 30
• for n = 3, f(n) = 100
• for n = 4, f(n) = 354
• for n = 5, f(n) = 1300

Observing above periodicity we can see that if (n%4==0) result of f(n)%5 is going to be 4 other wise result = 0. So, rather than calculating actual value of f(n) and then obtaining its value with mod 5 we can easily get result only by examine value of n. 

C++

// Program to find value of f(n)%5
#include <bits/stdc++.h>
using namespace std;

// function for obtaining remainder
int fnMod(int n)
{
    // calculate res based on value of n
    return (n % 4) ? 0 : 4;
}

// driver program
int main()
{
    int n = 43;
    cout << fnMod(n) << endl;
    n = 44;
    cout << fnMod(n);
    return 0;
}

Java

// Program to find value of f(n)% 5

class GFG
{
    // function for obtaining remainder
    static int fnMod(int n)
    {
        // calculate res based on value of n
        return (n % 4 != 0) ? 0 : 4;
    }

    // Driver code
    public static void main (String[] args)
    {
        int n = 43;
        System.out.println(fnMod(n));
        n = 44;
        System.out.print(fnMod(n));
    }
}

// This code is contributed by Anant Agarwal.

Python

# program to find f(n) mod 5
def fnMod (n):
    res = 4 if (n % 4 == 0) else 0
    return res

# driver section
n = 43
print (fnMod(n))
n = 44
print (fnMod(n))

C#

// C# Program to find value of f(n) % 5
using System;

class GFG {

    // function for obtaining remainder
    static int fnMod(int n)
    {
        // calculate res based on value of n
        return (n % 4 != 0) ? 0 : 4;
    }

    // Driver code
    public static void Main ()
    {
        int n = 43;
        Console.WriteLine(fnMod(n));
        n = 44;
        Console.Write(fnMod(n));
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP Program to find value of f(n)%5

// function for obtaining remainder
function fnMod($n)
{

    // calculate res based
    // on value of n
    return ($n % 4) ? 0 : 4;
}

// Driver Code
{
    $n = 43;
    echo fnMod($n),"\n" ;
    $n = 44;
    echo fnMod($n);
    return 0;
}

// This code is contributed by nitin mittal.
?>

Javascript

<script>
// JavaScript program to find value of f(n)% 5

// function for obtaining remainder
    function fnMod(n)
    {
        // calculate res based on value of n
        return (n % 4 != 0) ? 0 : 4;
    }


// Driver Code

        let n = 43;
        document.write(fnMod(n) + "<br/>");
        n = 44;
        document.write(fnMod(n)  + "<br/>");

// This code is contributed by splevel62.
</script>

Output:

0
4

Time complexity: O(1)
Auxiliary space: O(1)