# Modular Arithmetic

Last Updated : 09 Apr, 2024

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 a 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:Â

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:

 int modpower(int x, int n, int m)Â {Â  Â if (n == 0)Â Â  Â  Â  Â return 1%m;Â  Â long long u = modpower(x,n/2,m); Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Â  Â u = (u*u)%m;Â  Â if (n%2 == 1)Â Â  Â  Â  Â u = (u*x)%m;Â  Â return u;}
C++ ```#include<bits/stdc++.h> #include<iostream> 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<<modpower(5,2,7)<<endl; return 0; } ``` C ```#include <stdio.h> //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

Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  xmâˆ’1 mod m = 1Â

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

Below are some more important concepts related to Modular ArithmeticÂ

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