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 Bint 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) % 5int 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 codeint 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 Bdef 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) % 5def 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 codeN = "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 Bfunction 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 % $a;}// Function to return // (1^n + 2^n + 3^n + 4^n) % 5function findMod($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 Codevar N = "4"; // Function calldocument.write(findMod(N)); // This code is contributed by Kirti</script> |
Output:
4
Time Complexity: O(|N|), where |N| is the length of the string.
Auxiliary Space: O(1), since no extra space has been taken.
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