Question

Find the number of matrixes a and b that on multiplication mod 2 gives c

Answer 1

Answer:

Modular multiplication. ... We will prove that (A * B) mod C = (A mod C * B mod C) mod C ... LHS = (C * C * Q1 * Q2 + C * Q1 * R2 + C * Q2 * R1 + R1 * R 2 ) mod C ..... You multiply 4 times 7 normally to get 28 and since one number is positive and one is negative the answer

Step-by-step explanation:

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

Answer 2

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How to calculate the modulo - an example

• Start by choosing the initial number (before performing the modulo operation). ...

• Choose the divisor. ...

• Divide one number by the other, rounding down: 250 / 24 = 10 . ...

• Multiply the divisor by the quotient. ...

• Subtract this number from your initial number (dividend