I am having a very difficult time trying to understand what exactly this question is asking.

How many valid English plaintexts are there for the ciphertext message CJU using a length-3, one-time pad of cyclic shifts, (i, j, k)?

If the encrypted question message is CJU and the one time pad is i, j, k would there not be exactly 1 solution? We are given the one time pad i, j, k so we can derive the plaintext simply by some alagebra

(x + 9) % 26 = 3

(y + 10) % 26 = 10

(z + 11) % 26 = 21

Where x, y, and z are the plaintext letters T, Z, J?

I feel like this is a trick question that I am not understanding

• I think you are misreading the statement, $(i, j, k)$ isn't supposed to be the key $(9, 10, 11)$, but some key. – Aristu Oct 16 '16 at 3:36
• how does i, j, k help me narrow down my resulting set? – TemporaryFix Oct 16 '16 at 3:39

$(i,j,k)$ is not a specific OPT. It's three variable names that you can use to talk about the key. The question is just asking, "Given all possible three-character OTPs, how many valid English plaintexts are there for the ciphertext CJU?"
In other words, how many different values are there for the triple $(i,j,k)$ such that "CJU" decrypts to an actual English word.
• ahhhh, so $(i, j, k)$ is just a generic. So the solution is every possible english 3 letter word since we do not know the pad – TemporaryFix Oct 16 '16 at 18:26