# Shots, shots, shots, shots!

Once in a while I go out for a drink to my favourite bar. Normally is just drink a beer, eat some snacks and have a good time with my three friends. But last night I wanted to do some shots. The bartender said there were only 4 shots available:

1-digit : contains 2 types of high-percentage liquors and 3 types of low-percentage liquors

2-digit : contains 2 types of high-percentage liquors and 0 types of low-percentage liquors

..-digit : contains...

5-digit : contains 1 type of high-percentage liquors and 6 types of low-percentage liquors

But the bartender forgot the name of the third shot and what it contains… He told me: find out what shot it is, and the next shots are on the house! For all four! plus those other two at the bar. Can you guys help me out and getting free shots for everyone at the bar?

Hint 1:

I asked the bartender for more information about the shots menu and he said: Actually there were 5 possible shots, but the fourth was taken of the menu because people said it tasted like it contained 0 types of high-percentage liquor and 0 types of low-percentage liquor. The fourth shot is called 4-digit.

Hint 2:

The bartender said that the order of mixing is important to him. When he knows the name of the shots (by counting to 5), he knows how many high-percentage liquors he has to pour first and secondly how many low-percentage liquors he has to pour. Thus for the 1-digit shot he pours first 2 and then 3, for the 2-digit: 2,0; 4-digit: 0,0 and 5-digit: 1,6. The bartender also found out that the amount of people he would have to give a free shot, is an important number!

Hint 3:

During deciphering the shot, we talked to the bartender. He just has a new hobby, encrypting messages. He mentioned he recently started, therefore only uses the basic method concerning letters and numbers. He is a more experienced mathematician and just had an exam last week. He knew for sure that he answered question 1a, 2b and 3c correctly by using multiplication, addition, brackets and subtraction to find the correct 2 digit number he could split. He told us this when he was preparing the two + four! free shots, because he thinks we can find the answer!

Hint 4:

$$2+4! = 2+4*3*2*1 = 26$$

Hint 5:

The solution can be found through dechipering the name of the shot by combining the number in the name of the shot, the word digit and mathematical operations. This will lead to a number which can be translated into a 2-digit number (23 for example). This only works one way, from the name of the shot to the 2-digit number.

Hint 6:

The barman confirms to us that the letters of the word digit can be transformed using A1Z26 ciphering. Using multiplication and addition depending on the number of the shot (alle 5 letters in digit are used) a number comes out. By using the amount of people who get a free shot the different types of high-and low-percentage liqours can be found (this system only works one way!).

• when mixing a shot, does the order matter? e.g. for a 1-digit, are the two high-percentage liquors added first, then the 3 low-percentage liquors, or is that not relevant? – Dirigible Mar 11 at 18:43
• Yes, shots are always made by pouring first the high-percentage liquors and then combining this with the low-percentage liquors. I added a new hint concerning this. – tyui Mar 12 at 8:05

2 types of high-percentage liquors and 1 types of low-percentage liquors?

Because

the word "digit" in A1Z26 is $$4, 9, 7, 9, 20$$

And the fact that

1-digit: $$4 + 9 + 7 + 9 + 20 = 49 \equiv 23 \mod 26$$
2-digit: $$4 \times 9 + 7 + 9 + 20 = 72 \equiv 20 \mod 26$$
3-digit: ???
4-digit: $$4 \times 9 \times 7 \times 9 + 20 = 2288 \equiv 0 \mod 26$$
5-digit: $$4 \times 9 \times 7 \times 9 \times 20 = 45360 \equiv 16 \mod 26$$

3-digit: $$4 \times 9 \times 7 + 9 + 20 = 281 \equiv 21 \mod 26$$

Which matches Hint 6 in that (Those not after "-" is the original text):

The barman confirms to us that the letters of the word digit can be transformed using A1Z26 ciphering.
- This is how we get the number $$4, 9, 7, 9, 20$$.
Using multiplication and addition depending on the number of the shot (alle 5 letters in digit are used) a number comes out.
- See the calculation above.
By using the amount of people who get a free shot the different types
- It uses mod 26, which is the amount of people who get a free shot
(this system only works one way!)
- The modulo function is many-to-one; you have no way to know the original number if you only know the result.

This is not a full answer, but it looks like you're getting quite desperate for some response to this, so I'll write up the thoughts I've had so far, in the hope that either someone can take it and continue to get the full answer, or at least that you will see which parts are clear to potential solvers and which parts may still need hints.

Basically we're looking for some kind of function $$f$$ such that $$f(1)=(2,3),f(2)=(2,0),f(4)=(0,0),f(5)=(1,6),$$ and it seems $$f(n)$$ may relate somehow to an $$n$$-digit number or numbers. Also this may only work for $$n=1,2,3,(4),5$$ as there is no mention of possible extensions to the menu.

We know that somehow

number-letter correspondences will be important. The number of people he'll offer a free drink is $$26$$.

Also

the order of high-percentage and low-percentage is important. That could be as simple as just using them for tens place and units place in a number ($$23,20,?,0,16$$ aka $$W,T,?,0,P$$), or it could be something else.

From the penultimate hint and the new tags, it seems the answer will be something to do with

putting numbers together to form other numbers, using $$\times,+,(,),-$$ and maybe $$!$$ as operations.

So how could this be?

• $$f(n)$$, as a two-digit number, is the number of ways to do something? Unlikely, with 23 possibilities for the 1-digit problem.

• $$f(n)$$, as two separate digits, describes some way of forming $$n$$-digit numbers? Like all $$n$$-digit numbers can be formed from those two digits using the given operations (that's not the answer, but something like that)?

• Maybe the numbers-to-letter conversion is for the left-hand side? So instead of considering $$n$$-digit numbers, we consider $$n$$-letter words. But again, 23 possibilities for anything involving 1-letter words (of which there are only two) seems unlikely.

• I'll happily delete this answer if people feel it's too partial, or not saying anything that wasn't obvious from the hints. I don't expect to get the bounty since surely somebody will get a fuller answer before then. – Rand al'Thor Mar 27 at 16:04
• I'm not sure this is much of an answer, as you acknowledge -- it seems to be more speculation than answer, as I don't think you say anything really concrete here. – Deusovi Mar 27 at 16:06
• Granted, but I can see the OP's frustration here: they're adding hint after hint but nobody is really responding. At least seeing somebody's thoughts and directions for an attempted solution might help them to decide which aspect to hint at next - or of course, as always, help someone else to get a fuller answer. – Rand al'Thor Mar 27 at 16:09
• True, thoughts like this could definitely be helpful to OP! But that doesn't make them an answer. I'd summarize them in comments, or maybe make a chat room for this puzzle and share thoughts there. – Deusovi Mar 27 at 16:16
• @Randal'Thor You're right about getting a bit desperate :). You're thoughts aren't that bad too. The number 26 is important in letter-number correspondences, but not in the way you describe it in you're second blockqoute (the found numbers are correct btw). Also the number 26 is used in another way than letter-number correspondences. – tyui Mar 27 at 17:27