Multiplicative Inverse Modulo Calculator

Find modular multiplicative inverse using Extended Euclidean Algorithm with step-by-step solutions

Calculate Multiplicative Inverse Modulo

The number whose multiplicative inverse we want to find

The modulus (must be positive)

Modular Multiplicative Inverse Result

No Multiplicative Inverse
Modulus must be positive

Why? The multiplicative inverse of a modulo m exists if and only if gcd(a, m) = 1. Since gcd(NaN, 0) = 0, the numbers are not coprime and the inverse doesn't exist.

Example Calculations

Basic Example

Find inverse of 3 modulo 7

a = 3, m = 7

Prime Modulus

Prime modulus ensures inverse exists

a = 5, m = 11

Larger Numbers

Modular inverse with larger numbers

a = 17, m = 43

No Inverse Case

Example where inverse doesn't exist

a = 6, m = 9

Key Concepts

Definition

x is the multiplicative inverse of a mod m if a × x ≡ 1 (mod m)

Existence Condition

Inverse exists ⟺ gcd(a, m) = 1 (coprime)

Prime Modulus

If m is prime, all non-zero a < m have inverses

Extended GCD

Most efficient algorithm for finding modular inverses

Algorithm Comparison

Extended Euclidean

O(log min(a,m)) - Very efficient

🐌

Brute Force

O(m) - Educational only

📚

Fermat's Little Theorem

a^(p-2) mod p when p is prime

Understanding Modular Multiplicative Inverse

What is Modular Multiplicative Inverse?

The modular multiplicative inverse of a modulo m is a number x such that:

a x x ≡ 1 (mod m)

This means when you multiply a and x, the remainder when divided by m is 1.

When Does It Exist?

  • Coprime condition: gcd(a, m) = 1
  • Prime modulus: If m is prime, inverse exists for all 1 ≤ a < m
  • Composite modulus: Only for numbers coprime to m

Extended Euclidean Algorithm

Step 1: Find GCD

Use Euclidean algorithm to find gcd(a, m)

Step 2: Bézout's Identity

Find x, y such that ax + my = gcd(a, m)

Step 3: Extract Solution

If gcd = 1, then x is our multiplicative inverse

Worked Example

Find inverse of 3 modulo 7:

7 = 2 x 3 + 1

3 = 3 x 1 + 0

gcd(3, 7) = 1 ✓

Working backwards:

1 = 7 - 2 x 3

So: 3 x (-2) + 7 x 1 = 1

Therefore: 3 x 5 ≡ 1 (mod 7)

Answer: 5

Applications

Cryptography

RSA encryption, Diffie-Hellman key exchange

Linear Congruences

Solving equations ax ≡ b (mod m)

Number Theory

Modular arithmetic, group theory

Computer Science

Hash functions, pseudo-random generators