All-Army CyberStakes 4 - Cryptography (Rotate Me)

This one is a very simple problem, especially considering it’s only worth 5 points. We start with a text file that contains enciphered text, which already seems to be in the expected format of a flag.

IKQ{KzGxBw_NcV_nWz_ItT_ScrflorN}

Based on the name of the challenge and the hints mentioning “Julius”, it’s safe to assume that this is a Caesar cipher, also known as a Rotation (ROT-N) cipher. This is a very simple enciphering method where you “shift” each character to the right by N number of places. Going beyond “Z” just causes you to rotate back around to “A”, which is where the term Rotation cipher comes from. In order to decrypt the text, you just need to shift the characters the same number of places in the other direction to the left. For example, in a classic Caesar (or ROT-3) cipher the encryption and decryption process looks like this:

Plaintext:  HELLO WORLD
Ciphertext: KHOOR ZRUOG

Since you rotate back to “A” when you hit “Z”, anything above a ROT-25 cipher is equivalent to a cipher between ROT-0 to ROT-25. (You can calculate exactly what it’s equivalent to by computing N mod 26) This means that the total number of possible keys in a Caesar cipher is only 25. This is easily brute-forced by hand and is exceptionally trivial for a computer to crack. However, if we wanted to be fancy about it, we can crack it like a known-plaintext attack. We know that the first three characters have to be ACI. By comparing first letters of the plaintext and ciphertext, we can calculate that the key is 9 - 1 = 8, since A is the 1st character of the alphabet and I is the 9th. So by rotating to the left by 8 spaces (or 18 spaces to the right, since -8 mod 26 = 18), we can decrypt the flag. It’s easy to do this using a common tool such as CyberChef.

Ciphertext: IKQ{KzGxBw_NcV_nWz_ItT_ScrflorN}
Plaintext:  ACI{CrYpTo_FuN_fOr_AlL_KujxdgjF}