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

### 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