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Find (1^n + 2^n + 3^n + 4^n) mod 5 | Set 2
  • Last Updated : 11 Jun, 2021

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

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

 

Approach: (1n + 2n + 3n + 4n) mod 5 = (1n mod ?(5) + 2n mod ?(5) + 3n mod ?(5) + 4n 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 = (1n mod 4 + 2n mod 4 + 3n mod 4 + 4n 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#




// 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>
Output: 
4

 

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