Modular Arithmetic

Modular arithmetic is a system of arithmetic for numbers where numbers "wrap around" after reaching a certain value, called the modulus. It mainly uses remainders to get the value after wrap around. It is often referred to as "clock arithmetic. As you can see, the time values wrap after reaching 12 and 9 + 4 = 13 is computed as remainder of 13 when divided by 12. The concept is widely used in various fields, including cryptography, computer science, and engineering.

When dividing two integers, we get an equation like this:
\dfrac{A}{B} = Q \text{ remainder } R

Where:

• A is the dividend
• Q is the quotient
• B is the divisor
• R is the remainder

The modulo operator (mod) helps us focus on the remainder:
A mod B=R

Example:

\frac{13}{5}=2 remainder 3
13mod  5=3

So, dividing 13 by 5 gives a remainder of 3

What is Modular Arithmetic?

In modular arithmetic, numbers are reduced within a certain range, defined by the modulus. For two integers a and b, and a positive integer n, we say that a is congruent to b modulo n if their difference is an integer multiple of n. This is denoted as:

a ≡ b (mod n)

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, meaning:
20 = 6 x 3 + 2

Read more about Quotient Remainder Theorem.

Modular Operations

Modular Addition

The rule for Modular Addition is:

(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 Divisionis 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: 

It is important that in the case of an even n, the value of x^n/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.

#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;
}

#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;
}

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));
    }
}

#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))

// 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.

Fermat’s theorem states that

                                              x^m−1 mod m = 1 

when m is prime and x and m are coprime. This also yields

x^k mod m = x^{k mod (m−1)} mod m.

Applications of Modular Arithmetic

1. Cryptography: Modular arithmetic is fundamental in cryptography, particularly in public-key cryptosystems like RSA, which relies on the difficulty of factoring large numbers and properties of modular exponentiation.

2. Computer Science: Modular arithmetic is used in hashing algorithms, checksums, and cryptographic hash functions to ensure data integrity and security.

3. Number Theory: In number theory, modular arithmetic helps solve congruences and Diophantine equations, contributing to the understanding of integer properties and relationships.

4. Digital Signal Processing: Modular arithmetic is used in algorithms for efficient computation in digital signal processing, particularly in the Fast Fourier Transform (FFT) and error-correcting codes.

5. Clock Arithmetic: The concept of modular arithmetic is akin to how clocks work, where the hours wrap around after reaching 12 or 24.

Solved Examples of Modular Arithmetic

Example 1: Problem: Show that 38 ≡ 14 (mod 12)

Solution:

38 = 3 × 12 + 2
14 = 1 × 12 + 2

Both 38 and 14 have the same remainder (2) when divided by 12.

Therefore, 38 ≡ 14 (mod 12)

Example 2: Addition in Modular Arithmetic

Problem: Compute (27 + 19) mod 7

Solution:

27 ≡ 6 (mod 7) because 27 = 3 × 7 + 6
19 ≡ 5 (mod 7) because 19 = 2 × 7 + 5
(27 + 19) mod 7 ≡ (6 + 5) mod 7 ≡ 11 mod 7 ≡ 4

Example 3: Multiplication in Modular Arithmetic

Problem: Compute (23 × 17) mod 5

Solution:

23 ≡ 3 (mod 5) because 23 = 4 × 5 + 3
17 ≡ 2 (mod 5) because 17 = 3 × 5 + 2
(23 × 17) mod 5 ≡ (3 × 2) mod 5 ≡ 6 mod 5 ≡ 1

Example 4: Modular Exponentiation

Problem: Compute 7^100 mod 11

Solution:

Use Euler's theorem: a^φ(n) ≡ 1 (mod n) for coprime a and n

φ(11) = 10 (Euler's totient function for prime 11)
7^100 ≡ 7^(10 × 10) ≡ (7^10)^10 ≡ 1^10 ≡ 1 (mod 11)

Example 5: Linear Congruence

Problem: Solve 5x ≡ 3 (mod 7)

Solution:

Multiply both sides by 3 (modular multiplicative inverse of 5 mod 7):

3 × 5x ≡ 3 × 3 (mod 7)
15x ≡ 9 (mod 7)
x ≡ 2 (mod 7)

Example 6: Chinese Remainder Theorem

Problem: Solve the system of congruences:
x ≡ 2 (mod 3)
x ≡ 3 (mod 5)
x ≡ 2 (mod 7)

Solution:

M = 3 × 5 × 7 = 105

M1 = 105/3 = 35, y1 = 35^(-1) mod 3 = 2
M2 = 105/5 = 21, y2 = 21^(-1) mod 5 = 1
M3 = 105/7 = 15, y3 = 15^(-1) mod 7 = 1

x = (2 × 35 × 2 + 3 × 21 × 1 + 2 × 15 × 1) mod 105
= (140 + 63 + 30) mod 105
= 233 mod 105
= 23

Verify: 23 ≡ 2 (mod 3), 23 ≡ 3 (mod 5), 23 ≡ 2 (mod 7)

Example 7: Modular Inverse

Problem: Find the modular inverse of 3 modulo 11

Solution:

Use extended Euclidean algorithm:

11 = 3 × 3 + 2
3 = 1 × 2 + 1
2 = 2 × 1 + 0

Working backwards:

1 = 3 - 1 × 2
1 = 3 - 1 × (11 - 3 × 3)
1 = 4 × 3 - 1 × 11

Therefore, 3^(-1) ≡ 4 (mod 11)
Verify: (3 × 4) mod 11 = 12 mod 11 = 1

Example 8: Fermat's Little Theorem

Problem: Calculate 7^222 mod 11

Solution:

Fermat's Little Theorem states that if p is prime and a is not divisible by p, then:
a^(p-1) ≡ 1 (mod p)

Here, p = 11 (prime) and a = 7 (not divisible by 11)
So, 7^10 ≡ 1 (mod 11)

Now, 222 = 22 × 10 + 2
Therefore, 7^222 ≡ (7^10)^22 × 7^2 (mod 11)
≡ 1^22 × 7^2 (mod 11)
≡ 49 (mod 11)
≡ 5 (mod 11)

Practice Problems on Modular Arithmetic

1. Calculate 47 mod 7.

2. Solve the linear congruence: 5x ≡ 3 (mod 8)

3. Find the modular multiplicative inverse of 5 modulo 11.

4. Using Fermat's Little Theorem, calculate 7¹⁰⁰ mod 11.

5. Solve the system of congruences: x ≡ 2 (mod 3)
x ≡ 3 (mod 5)
x ≡ 4 (mod 7)

6. Determine whether 29 is prime using Wilson's Theorem.

7. Calculate (17 × 23 + 31) mod 13.

8. Find the remainder when 3²⁰⁰ is divided by 7.

9. Solve the congruence: x² ≡ 4 (mod 11)

10. Given that 7¹⁰ ≡ 1 (mod 11), find 7¹⁰³ mod 11 without direct computation.

Related Articles:

• Euler’s Totient Function
• Compute n! under modulo p
• Wilson’s Theorem
• How to compute mod of a big number?
• Find value of y mod (2 raised to power x)