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

Last Updated : 11 Jul, 2025

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 = (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

JavaScript

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Time Complexity: O(|N|), where |N| is the length of the string.
Auxiliary Space: O(1), since no extra space has been taken.