John's new password is:
The word keeps in the "money" theme. It is for a bank account (money-themed) and his previous password was BROKE (money-themed).
John's actions were as follows:
Alternate alternate solution:
I think the answer is
My password is 2
It is ten too.
My password is seven letters long
four letters is also not wrong
It is fair to say that is is none
but that could be off by one
If you think my password is a single hole,
I will blow up the north Pole!
The number you should say is:
We can decode each phrase by ..
There's a special rule, though. The name of the café hints at ...
With this method, the first phrase becomes:
The last hint:
So, now to my passphrase:
Jeez, I finally got it (well, at least I hope so).
The logic for admission is a classic one, but with a silly twist:
This doesn't seem to work exactly right, so here's the twist:
So this is what happens:
About the high-five guy:
Finally, the correct answer for the author:
For the known-good answers we have so far [EDITED to add: ... but more answers have since been added to the question], the number equals
with the proviso that
In that case, depending on whether we take the cashier's last utterance as a "prompt" or an "error message", it seems like you need to say
Solving Key 2 as a keyed Caesar cipher with the key being the answer from room 1, you get:
The H's in the string are incorrectly coded to F's - they should be Y's in the coded version but they are presented as U's.
H being coded to F in the string could mean we should also code it to F in the chemical (C12F27N instead of C12H27N)
Then, the sum ...
Since this puzzle has so many facets, I've decided to try a sort of "community answer", a compilation of solvers' efforts. If you make a chunk of progress, simply edit this post and add it. This way the final result will be organized in a clear manner that is great for progeny's sake. Thanks and have fun!
1 Over thirty years old and still I dig;...
The angels should play the game, as long as the formula for the computable function must be fixed in advance and finite.
Proof: let F be the shortest encoding of the formula (length l(F)) using a set of symbols S of size n(S). Then the guessing angel can simply count up from 0 in base n(S), and some time before n(S)l(F) they will encounter the formula.
Take the given routes as vectors and sum them, then apply the resulting vector to Aberdeen. Using latitude and longitude isn't quite accurate because of the curvature of the earth, but it's close enough.
Tehran =35°42′N 51°25′E
Brussels to Tehran: -15°09′ +47°04′
Warsaw =52°14′N 21°01′E
Warsaw to ...