# Modular multiplication

## Let's explore the multiplication property of modular arithmetic:

(A * B) mod C = (A mod C * B mod C) mod C

### Example for Multiplication:

Let A=4, B=7, C=6
Let's verify: (A * B) mod C = (A mod C * B mod C) mod C
LHS= Left Hand Side of the Equation
RHS= Right Hand Side of the Equation
LHS = (A * B) mod C
LHS = (4 * 7) mod 6
LHS = 28 mod 6
LHS = 4
RHS = (A mod C * B mod C) mod C
RHS = (4 mod 6 * 7 mod 6) mod 6
RHS = (4 * 1) mod 6
RHS = 4 mod 6
RHS = 4
LHS = RHS = 4

## Proof for Modular Multiplication

We will prove that (A * B) mod C = (A mod C * B mod C) mod C
We must show that LHS = RHS
From the quotient remainder theorem we can write A and B as:
A = C * Q1 + R1 where 0 ≤ R1 < C and Q1 is some integer. A mod C = R1
B = C * Q2 + R2 where 0 ≤ R2 < C and Q2 is some integer. B mod C = R2
LHS = (A * B) mod C
LHS = ((C * Q1 + R1 ) * (C * Q2 + R2) ) mod C
LHS = (C * C * Q1 * Q2 + C * Q1 * R2 + C * Q2 * R1 + R1 * R 2 )  mod C
LHS = (C * (C * Q1 * Q2 + Q1 * R2 + Q2 * R1)  + R1 * R 2 )  mod C
We can eliminate the multiples of C when we take the mod C
LHS = (R1 * R2) mod C
Next let's do the RHS
RHS = (A mod C * B mod C) mod C
RHS = (R1 * R2 ) mod C
Therefore RHS = LHS
LHS = RHS = (R1 * R2 ) mod C