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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 ``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

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## Javascript

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Output:
`4`

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