Cheryl chooses a word in this problem and tells its first letter to Aerith and its last letter to Bob, The following conversation ensues over a series of emails:

Aerith: "I don't know her word, do you?"

Bob: "No, in fact, I don't know if we can ever figure out what her word is without having more information."

Aerith: "Then I do know what it is!"

Bob: "Now I also do."[^0]

What is Cheryl's word?
We will process their conversation message by message.

"I don't know her word, do you?"

Aerith would know Cheryl's word if the first letter was unique, so Aerith does not have any of b (Bob), $\mathrm{k}$ (know), $\mathrm{m}$ (more), $\mathrm{p}$ (problem), s (series), y (you).

"No, in fact, I don't know if we can ever figure out what her word is without having more information."

If they can't ever figure out what Cheryl's word is, even if they just outright told each other their letters, it would be because Cheryl's word is not identifiable from these letters alone. The unidentifiable words are: can/conversation, Cheryl's/chooses, emails/ensues, fact/first, in/information, is/its, tells/this, what/without.

Bob's statement is then equivalent to the claim "The word could be in this list but does not have to be." The only last letters which satisfy this property are $\mathrm{n}$ and $\mathrm{t}$. While some words in the list end in s, s does not satisfy the property because no words not in the list end in s. (Series was the only one but it was eliminated by Aerith's statement.)

## "Then I do know what it is!"

In order for Aerith to make this claim, there must be only one possible word beginning with her letter that ends with $\mathrm{n}$ or $\mathrm{t}$. Her letter must then be one of $\mathrm{d}$ (don't), $l$ (last), o (out), $\mathrm{t}$ (then).

## "Now I also do."

For Bob to know Cheryl's word, the last letter must be unique among the remaining options. Inspecting we see that the word must then be "then".