Reverse – Vault door 6

Vault door 6 - 350pt

Challenge

This vault uses an XOR encryption scheme. The source code for this vault is here: VaultDoor6.java

Hints

If X ^ Y = Z, then Z ^ Y = X. Write a program that decrypts the flag based on this fact.

Solution

Again, there’s a list of numbers meant to be read as ASCII codes. The only difference from Level 4 is that you have to xor every number with 0x55 before decoding.

picoCTF{n0t_mUcH_h4rD3r_tH4n_x0r_aedeced}

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