Question

2.Consider the following credit card numbers and tell me if they are valid or not. If a number is invalid, write the valid credit number by changing the “check” digit. Show your work for full marks. a. 4519 6731 6055 4166

b. 4515 1663 4722 3333

c. How many valid (according to the Luhn check) 16-digit credit card numbers are there that start with these 14 digits 4444 4444 4444 45? Give one of the valid credit card numbers.

3. We have the following encryption algorithm: XOR with the HEX key AA, rotate 4 places to the right, and XOR with the HEX key BC. Show your work for full marks.

a. If we start with the bit pattern represented by the HEX number ACE1, what would the HEX number be that represents the encrypted bit pattern?

b. Assume that we have intercepted a message that was encoded using the same algorithm. The encoded message as a HEX number was D007. What was the original message as a HEX number?

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Answer 1

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Answer ::

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b) 4515 1663 47 22 3333 odd digit > Step 1: original number 45 ir 1663 4722 3383 steps: Drop last ligit 4.5 ir 166 347 2.2 3 3 3 Step 3: Reverse the digit 3 3 3 2 2 7 4 366 15154 Step 4: Muntiply 2 by 6 6 6 22 14 4 6 6 6 1 10 1 104 Stop risustract g to numbers overs. 66 62 25 4 6 6 61 steps: Add all digit 57 step7: mod 10: 57% mod 10=7 Last digit 7 7 sogits not valid card nurser x x
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3) Encryption algorithm

XOR with the HEX key BC

rotate 4 places to the left

XOR with the HEX key AA

Decryption algorithm

XOR with the HEX key AA

rotate 4 places to the right

XOR with the HEX key BC

a. If we start with the bit pattern represented by the HEX number ACE1, what would the HEX number be that represents the encrypted bit pattern?

Plaintext = ACE1

Encryption

Step 1: ACE1 XOR BCBC (Key is repeated twice to match the length)

= 1010 1100 1110 0001 XOR 1011 1011 1100 1100

= 0001 0111 0010 1101 => 172D

Step 2: Rotate 0001 0111 0010 1101 4 places left

= 0111 0010 1101 0001 => 72D1

Step 3: 72D1 XOR AAAA (Key is repeated twice to match the length)

= 0111 0010 1101 0001 XOR 1010 1010 1010 1010

= 1101 1000 0111 1011 => D87B

Ciphertext = D87B

b. Assume that we have intercepted a message that was encoded using the same algorithm. The encoded message as a HEX number was D007. What was the original message as a HEX number?

Ciphertext = D007

Decryption

Step 1: D007 XOR AAAA (Key is repeated twice to match the length)

= 1101 0000 0000 0111 XOR 1010 1010 1010 1010

= 0111 1010 1010 1101 => 7AAD

Step 2: Rotate 0111 1010 1010 1101 4 places right

= 1101 0111 1010 1010 => D7AA

Step 3: D7AA XOR BCBC (Key is repeated twice to match the length)

= 1101 0111 1010 1010 XOR 1011 1011 1100 1100

= 0110 1100 0110 0110 => 6C66

Plaintext = 6C66