The associated script, ciphertext, and PDF of the solution can be found in this repository:
Script, ciphertext, and PDFFor this challenge, I inferred from the name that this is a single byte xor challenge. This means that every byte of the plaintext was XOR’ed with a single byte key and the output was the ciphertext. I know that if I find the encryption key and XOR the ciphertext with the encryption key, I will get back the plaintext due to the following property: A XOR B = C => C XOR B = A and C XOR A = B. Thus, I created a python script (sbxor.py) that XOR’ed the encrypted text with every possible value for a byte. Eventually, I will have to have encountered the key. Then, I inspected my output and obtained the flag.