Actually you maybe don't even need the key. I don't know what algorithm Adobe used to encrypt these passwords, but for most standard ones it holds that if you have two encrypted messages A' and B' corresponding to plain texts A and B, then B = A ^ A' ^ B'. You really only need to find one password of length N to be able to decrypt all passwords with length <= N.

Edit: Of course I mean two messages encrypted with the same key and initialisation vector. Do not reuse IVs kids, you'll get hurt.