8-bit CFB in DES and AES - error propagation
17 bytes are affected in AES mode, not 17 bits. This is because the block size of AES is 16 bytes instead of 8 bytes for DES. 

When you change the ciphertext it directly affects the bits that are at the same location of the plaintext, plus the time it takes to get out of the shift register, which is the size of the block size of the cryptographic primitive, i.e. 8 bytes for DES and 16 bytes for AES.

The question is however how many blocks are affected, not how many bytes. Note that the block size of CFB will be set to 1 bit, not the block size of the underlying block cipher. CFB is a stream mode after all.

So you could say that 9 * 8 = 72 blocks are affected for DES or indeed 17 * 8 = 136 for AES. This is probably not what is meant by the teacher, you're better off guessing 2 blocks for both or 9 for DES and 17 for AES. Also see the explanation by SEJPM here.



The nice thing about CFB8 mode is that if you drop or insert a byte that it will re-synchronize. What it won't do is to only flip those plaintext bits that are changed in the ciphertext, such as CTR mode does.

Nowadays we're not that interested anymore. We tend to catch transmission errors in a lower layer and then only decrypt if the entire ciphertext is available, preferably using authenticated encryption to catch any changes to the ciphertext (deliberate or otherwise).

Note also that CFB8 mode is terribly expensive; it requires a block encrypt for each byte that is transmitted, both for encrypting as decrypting.