There's a cool new cybercafé in town called #DECAFE
The café charges a fee to use their computers, but if you BYOC (bring your own computer) you can get on their internet free with any purchase... if you know how!
After you pay, the cashier says a sentence. If you reply with the correct number, they will hand you a slip with credentials.
You eavesdrop as the four people in front of you acquire internet credentials:
Cashier: Aloof whale swim forth.
Customer #1: One
Cashier: Tasty warm pasties contain stuff.
Customer #2: Two
Cashier: Pinkish fish adds distinct flavour on your bagel: robust smack.
Customer #3: Three
At this point, you figure it's easy enough, but then you're thrown for a loop when the next number is not four.
Cashier: Toxic frogs dispatch.
Customer #4: Sixty
After you've made your purchase, the cashier says:
Our cat can lick ice, you know.
What number must you say to obtain internet credentials?
Clarification
The challenge-response format is just intended to be a fun wrapper (and to make solving the puzzle easier by providing explicit known hidden messages): the challenges conceal the appropriate responses plainly using steganographic methods.
Hints
#DECAFE
is a color as pointed out in the comments, but that's not what counts.
Every word is important, but only a handful of letters are meaningful.
The spelling of a number (in English) is relevant in at least one challenge. Not all numbers are represented the same way.
The cat could also pour ale without changing your answer. Similarly, the pasties could also contain pilaf, but not meats. Moreover, the whale could instead be a goose or a drake or possibly a manatee, but not an otter or a squid.
This phrasing with cryptical prose forms hint
#DECAFE
is a colour. $\endgroup$SIXTY
and60
are both possible as is at least one other representation.60
would not be6D
though that might be a fair representation for109
, however I will say none of the challenges in the question use hex in that manner.6T
is right out. $\endgroup$60
would be unlikely (not impossible, but I intentionally avoided something that could be used to make it possible to keep the puzzle more straightforward) for a reason that will be quite obvious once this is solved. $\endgroup$