Question

Assuming you are using a 64-bit (8-byte) block cipher in CBC mode. Show the results of padding the following plaintexts so they can fit into an even number of blocks.

(A) 0f8353c2 8e1ff6

(B) 8c521e

(C) c1

(D) 9f6bef2c 902fe4ba

Answer 1

We will use PKCS5 padding. It is most common type of padding.

In PKCS5, the padding bytes have thesame value and is equal to the number of padding bytes

I will write the padded bytes in bold

(A) 0f8353c2 8e1ff6

Here we need 1 padding byte, and the value will be 01

It is given that block size = 8 bytes

Block 1 = 0f8353c2

Block 2 = 8e1ff601

(B) 8c521e

Here we need 5 padding byte, and the value will be 05

Block 1 = 8c521e05

Block 2 = 05050505

(C) c1

Here we need 7 padding bytes, so the value will be 07

Block 1 = c1070707

Block 2 = 07070707

(D) 9f6bef2c 902fe4ba

Here we don't need any padding bytes

Block 1 = 9f6bef2c

Block 2 = 902fe4ba