Find (1^n + 2^n + 3^n + 4^n) mod 5 | Set 2

Last Updated : 09 Jun, 2022

Given a very large number N. The task is to find (1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ) mod 5.
Examples:

Input: N = 4
Output: 4
(1 + 16 + 81 + 256) % 5 = 354 % 5 = 4
Input: N = 7823462937826332873467731
Output: 0

Approach: (1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ) mod 5 = (1^{n mod ?(5)} + 2^{n mod ?(5)} + 3^{n mod ?(5)} + 4^{n mod ?(5)}) mod 5.
This formula is correct because 5 is a prime number and it is coprime with 1, 2, 3, 4.
Know about ?(n) and modulo of large number
?(5) = 4, hence (1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ) mod 5 = (1^{n mod 4} + 2^{n mod 4} + 3^{n mod 4} + 4^{n mod 4}) mod 5
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return A mod B
int A_mod_B(string N, int a)
{
    // length of the string
    int len = N.size();

    // to store required answer
    int ans = 0;
    for (int i = 0; i < len; i++)
        ans = (ans * 10 + (int)N[i] - '0') % a;

    return ans % a;
}

// Function to return (1^n + 2^n + 3^n + 4^n) % 5
int findMod(string N)
{
    // ?(5) = 4
    int mod = A_mod_B(N, 4);

    int ans = (1 + pow(2, mod) + pow(3, mod)
               + pow(4, mod));

    return (ans % 5);
}

// Driver code
int main()
{
    string N = "4";
    cout << findMod(N);

    return 0;
}

Java

// Java implementation of the approach 
class GFG
{
    
// Function to return A mod B 
static int A_mod_B(String N, int a) 
{ 
    // length of the string 
    int len = N.length(); 

    // to store required answer 
    int ans = 0; 
    for (int i = 0; i < len; i++) 
        ans = (ans * 10 + (int)N.charAt(i) - '0') % a; 

    return ans % a; 
} 

// Function to return (1^n + 2^n + 3^n + 4^n) % 5 
static int findMod(String N) 
{ 
    // ?(5) = 4 
    int mod = A_mod_B(N, 4); 

    int ans = (1 + (int)Math.pow(2, mod) + 
                (int)Math.pow(3, mod) + 
                (int)Math.pow(4, mod)); 

    return (ans % 5); 
} 

// Driver code 
public static void main(String args[]) 
{ 
    String N = "4"; 
    System.out.println(findMod(N)); 
}
} 

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach

# Function to return A mod B
def A_mod_B(N, a):
    
    # length of the string
    Len = len(N)

    # to store required answer
    ans = 0
    for i in range(Len):
        ans = (ans * 10 + int(N[i])) % a

    return ans % a

# Function to return (1^n + 2^n + 3^n + 4^n) % 5
def findMod(N):

    # ?(5) = 4
    mod = A_mod_B(N, 4)

    ans = (1 + pow(2, mod) + 
               pow(3, mod) + pow(4, mod))

    return ans % 5

# Driver code
N = "4"
print(findMod(N))

# This code is contributed by mohit kumar

// C# implementation of the approach 
using System;

class GFG
{
    
// Function to return A mod B 
static int A_mod_B(string N, int a) 
{ 
    // length of the string 
    int len = N.Length; 

    // to store required answer 
    int ans = 0; 
    for (int i = 0; i < len; i++) 
        ans = (ans * 10 + (int)N[i] - '0') % a; 

    return ans % a; 
} 

// Function to return (1^n + 2^n + 3^n + 4^n) % 5 
static int findMod(string N) 
{ 
    // ?(5) = 4 
    int mod = A_mod_B(N, 4); 

    int ans = (1 + (int)Math.Pow(2, mod) + 
                (int)Math.Pow(3, mod) + 
                (int)Math.Pow(4, mod)); 

    return (ans % 5); 
} 

// Driver code 
public static void Main() 
{ 
    string N = "4"; 
    Console.WriteLine(findMod(N)); 
}
} 

// This code is contributed by Code_Mech.

PHP

<?php
// PHP implementation of the approach

// Function to return A mod B
function A_mod_B($N, $a)
{
    // length of the string
    $len = strlen($N);

    // to store required answer
    $ans = 0;
    for ($i = 0; $i < $len; $i++)
        $ans = ($ans * 10 + 
               (int)$N[$i] - '0') % $a;

    return $ans % $a;
}

// Function to return 
// (1^n + 2^n + 3^n + 4^n) % 5
function findMod($N)
{
    // ?(5) = 4
    $mod = A_mod_B($N, 4);

    $ans = (1 + pow(2, $mod) + 
                pow(3, $mod) + pow(4, $mod));

    return ($ans % 5);
}

// Driver code
$N = "4";
echo findMod($N);

// This code is contributed 
// by Akanksha Rai
?>

JavaScript

<script>

// Javascript implementation of the approach 

// Function to return A mod B 
function A_mod_B(N, a) 
{ 
    // length of the string 
    var len = N.length; 

    // to store required answer 
    var ans = 0; 
    for (var i = 0; i < len; i++) 
        ans = (ans * 10 + parseInt(N.charAt(i) - '0')) % a; 

    return ans % a; 
} 

// Function to return (1^n + 2^n + 3^n + 4^n) % 5 
function findMod(N) 
{ 
    // ?(5) = 4 
    var mod = A_mod_B(N, 4); 

    var ans = (1 + parseInt(Math.pow(2, mod) + 
                Math.pow(3, mod) + 
                Math.pow(4, mod))); 

    return (ans % 5); 
} 

// Driver Code
var N = "4"; 
    
// Function call
document.write(findMod(N)); 

// This code is contributed by Kirti

</script>