Modular Arithmetic

Modular arithmetic is the branch of arithmetic mathematics related with the “mod” functionality. Basically, modular arithmetic is related with computation of “mod” of expressions. Expressions may have digits and computational symbols of addition, subtraction, multiplication, division or any other. Here we will discuss briefly about all modular arithmetic operations.

Quotient Remainder Theorem:

It states that, for any pair of integers a and b (b is positive), there exist two unique integers q and r such that:

a = b x q + r where 0 <= r < b

Example: If a = 20, b = 6 then q = 3, r = 2 20 = 6 x 3 + 2

(a + b) mod m = ((a mod m) + (b mod m)) mod m

Example:

```(15 + 17) % 7
= ((15 % 7) + (17 % 7)) % 7
= (1 + 3) % 7
= 4 % 7
= 4```

The same rule is to modular subtraction. We don’t require much modular subtraction but it can also be done in the same way.

Modular Multiplication:

The Rule for modular multiplication is:

(a x b) mod m = ((a mod m) x (b mod m)) mod m

Example:

```(12 x 13) % 5
= ((12 % 5) x (13 % 5)) % 5
= (2 x 3) % 5
= 6 % 5
= 1```

Modular Division:

The modular division is totally different from modular addition, subtraction and multiplication. It also does not exist always.

`(a / b) mod m is not equal to ((a mod m) / (b mod m)) mod m.`

This is calculated using the following formula:

(a / b) mod m = (a x (inverse of b if exists)) mod m

Modular Inverse:

The modular inverse of a mod m exists only if a and m are relatively prime i.e. gcd(a, m) = 1. Hence, for finding the inverse of an under modulo m, if (a x b) mod m = 1 then b is the modular inverse of a.

Example: a = 5, m = 7 (5 x 3) % 7 = 1 hence, 3 is modulo inverse of 5 under 7.

Modular Exponentiation:

Finding a^b mod m is the modular exponentiation. There are two approaches for this – recursive and iterative. Example:

```a = 5, b = 2, m = 7
(5 ^ 2) % 7 = 25 % 7 = 4```

There is often a need to efficiently calculate the value of xn mod m. This can be done in O(logn) time using the following recursion:

It is important that in the case of an even n, the value of xn/2 is calculated only once.

This guarantees that the time complexity of the algorithm is O(logn) because n is always halved when it is even.

The following function calculates the value of xn mod m:

C++

 `#include``#include` `using` `namespace` `std;` `//function that calculate modular exponentiation x^n mod m.``int` `modpower(``int` `x, ``int` `n, ``int` `m) ``{``    ``if` `(n == 0) ``//base case ``        ``return` `1%m; ``    ``long` `long` `u = modpower(x,n/2,m);  ``    ``u = (u*u)%m;``    ``if` `(n%2 == 1) ``//when 'n' is odd``        ``u = (u*x)%m;``    ``return` `u;``}` `//driver function``int` `main()``{ ``    ``cout<

C

 `#include ` `//function that calculate modular exponentiation x^n mod m.``int` `modpower(``int` `x, ``int` `n, ``int` `m) ``{``    ``if` `(n == 0) ``//base case ``        ``return` `1 % m; ``    ``long` `long` `u = modpower(x, n / 2, m);  ``    ``u = (u * u) % m;``    ``if` `(n % 2 == 1) ``// when 'n' is odd``        ``u = (u * x) % m;``    ``return` `u;``}` `//driver function``int` `main()``{ ``    ``printf``(``"%d\n"``, modpower(5, 2, 7));``    ``return` `0;``}`

Java

 `import` `java.util.*;` `class` `GFG {``    ``//function that calculate modular exponentiation x^n mod m.``    ``public` `static` `int` `modpower(``int` `x, ``int` `n, ``int` `m) {``        ``if` `(n == ``0``) ``//base case ``            ``return` `1` `% m; ``        ``long` `u = modpower(x, n / ``2``, m);  ``        ``u = (u * u) % m;``        ``if` `(n % ``2` `== ``1``) ``// when 'n' is odd``            ``u = (u * x) % m;``        ``return` `(``int``)u;``    ``}` `    ``//driver function``    ``public` `static` `void` `main(String[] args) {``        ``System.out.println(modpower(``5``, ``2``, ``7``));``    ``}``}`

Python

 `#function that calculate modular exponentiation x^n mod m.``def` `modpower(x, n, m):``    ``if` `n ``=``=` `0``: ``# base case``        ``return` `1` `%` `m``    ``u ``=` `modpower(x, n ``/``/` `2``, m)``    ``u ``=` `(u ``*` `u) ``%` `m``    ``if` `n ``%` `2` `=``=` `1``: ``# when 'n' is odd``        ``u ``=` `(u ``*` `x) ``%` `m``    ``return` `u` `#driver function``print``(modpower(``5``, ``2``, ``7``))`

Javascript

 `// function that calculates modular exponentiation x^n mod m.``function` `modpower(x, n, m) {``    ``if` `(n == 0) { ``// base case``        ``return` `1 % m;``    ``}``    ``let u = modpower(x, Math.floor(n / 2), m);``    ``u = (u * u) % m;``    ``if` `(n % 2 == 1) { ``// when 'n' is odd``        ``u = (u * x) % m;``    ``}``    ``return` `u;``}` `// driver function``console.log(modpower(5, 2, 7));`

```output:
4```

Time complexity: O(logn), because n is always halved when it is even.

Fermatâ€™s theorem states that

xmâˆ’1 mod m = 1

when m is prime and x and m are coprime. This also yields
xk mod m = xk mod (mâˆ’1) mod m.

Below are some more important concepts related to Modular Arithmetic

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