# Value of the series (1^3 + 2^3 + 3^3 + … + n^3) mod 4 for a given n

Given a function f(n) = (13 + 23 + 33 + … + n3), the task is to find the value of f(n) mod 4 for a given positive integer ‘n’.
Examples

```Input: n=6
Output: 1
Explanation: f(6) = 1+8+27+64+125+216=441
f(n) mod 4=441 mod 4 = 1

Input: n=4
Output: 0
Explanation: f(4)=1+8+27+64 = 100
f(n) mod 4 =100 mod 4 = 0
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

While solving this problem, you might directly compute (13 + 23 + 33 + … + n3) mod 4. But this requires O(n) time. We can use direct formula for sum of cubes, but for large ‘n’ we may get f(n) out of range of long long int.

Here’s an efficient O(1) solution:

From division algorithm we know that every integer can be expressed as either 4k, 4k+1, 4k+2 or 4k+3.
Now,

(4k)3 = 64k3 mod 4 = 0.
(4k+1)3 = (64k3 + 48k2 + 12k+1) mod 4 = 1
(4k+2)3 = (64k3+64k2+48k+8) mod 4 = 0
(4k+3)3 = (64k3+184k2 + 108k+27) mod 4 = 3
and ((4k)3+(4k+1)3+(4k+2)3+(4k+1)4) mod 4 = (0 + 1+ 0 + 3) mod 4 = 0 mod 4

Now let x be the greatest integer not greater than n divisible by 4. So we can easily see that,
(13+23+33…..+x3) mod 4=0.
Now,

• if n is a divisor of 4 then x=n and f(n) mod 4 =0.
• Else if n is of the form 4k + 1, then x= n-1. So, f(n)= 13 + 23 + 33…..+ x3+n3) mod 4 = n^3 mod 4 = 1
• Similarly if n is of the form 4k+2 (i.e x=n-2), we can easily show that f(n) = ((n-1)3+n3) mod 4=(1 + 0) mod 4 = 1
• And if n is of the form 4k+3 (x=n-3). Similarly, we get f(n) mod 4 = ((n-2)3+(n-1)3+n3) mod 4 = (1+0+3) mod 4 = 0

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// function for obtaining the value of f(n) mod 4 ` `int` `fnMod(``int` `n) ` `{ ` `    ``// Find the remainder of n when divided by 4 ` `    ``int` `rem = n % 4; ` ` `  `    ``// If n is of the form 4k or 4k+3 ` `    ``if` `(rem == 0 || rem == 3) ` `        ``return` `0; ` ` `  `    ``// If n is of the form 4k+1 or 4k+2 ` `    ``else` `if` `(rem == 1 || rem == 2) ` `        ``return` `1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 6; ` `    ``cout << fnMod(n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` ` `  `import` `java .io.*; ` ` `  `class` `GFG { ` `  `  `// function for obtaining the value of f(n) mod 4 ` `static` `int` `fnMod(``int` `n) ` `{ ` `    ``// Find the remainder of n when divided by 4 ` `    ``int` `rem = n % ``4``; ` ` `  `    ``// If n is of the form 4k or 4k+3 ` `    ``if` `(rem == ``0` `|| rem == ``3``) ` `        ``return` `0``; ` ` `  `    ``// If n is of the form 4k+1 or 4k+2 ` `    ``else` `if` `(rem == ``1` `|| rem == ``2``) ` `        ``return` `1``; ` `        ``return` `0``; ` `} ` ` `  `// Driver code ` ` `  `    ``public` `static` `void` `main (String[] args) { ` `            ``int` `n = ``6``; ` `    ``System.out.print( fnMod(n)); ` `    ``} ` `} ` `//This code is contributed  ` `// by shs `

## Python 3

 `# Python3 implementation of ` `# above approach ` ` `  `# function for obtaining the  ` `# value of f(n) mod 4  ` `def` `fnMod(n) : ` ` `  `    ``# Find the remainder of n ` `    ``# when divided by 4  ` `    ``rem ``=` `n ``%` `4` ` `  `    ``# If n is of the form 4k or 4k+3  ` `    ``if` `(rem ``=``=` `0` `or` `rem ``=``=` `3``) : ` `        ``return` `0` ` `  `    ``# If n is of the form  ` `    ``# 4k+1 or 4k+2  ` `    ``elif` `(rem ``=``=` `1` `or` `rem ``=``=` `2``) : ` `        ``return` `1` ` `  `# Driver code      ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``n ``=` `6` `    ``print``(fnMod(n)) ` ` `  `# This code is contributed  ` `# by ANKITRAI1 `

## C#

 `// C# implementation of the approach ` `  `  ` `  `using` `System;  ` `class` `GFG { ` `   `  `// function for obtaining the value of f(n) mod 4 ` `static` `int` `fnMod(``int` `n) ` `{ ` `    ``// Find the remainder of n when divided by 4 ` `    ``int` `rem = n % 4; ` `  `  `    ``// If n is of the form 4k or 4k+3 ` `    ``if` `(rem == 0 || rem == 3) ` `        ``return` `0; ` `  `  `    ``// If n is of the form 4k+1 or 4k+2 ` `    ``else` `if` `(rem == 1 || rem == 2) ` `        ``return` `1; ` `        ``return` `0; ` `} ` `  `  `// Driver code ` `  `  `    ``public` `static` `void` `Main () { ` `            ``int` `n = 6; ` `    ``Console.Write( fnMod(n)); ` `    ``} ` `} `

## PHP

 ` `

Output:

```1
```

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