Solution: Refreshment Counter

Answer: GORDITA

Upon initially opening this puzzle, solvers are presented with a banner saying “Congratulations! You are the (letters here)TH visitor to this website!”, some text describing the content below as “fun Fact or Fictions”, and a grid with 5 rows and 7 columns which has some squares filled with text.

Upon refreshing the page, some of the contents of the page will change. The banner on the top will have a different label of bolded letters, and the content filled in the grid below will likely be different. From refreshing the page several more times, the following can be deduced to a reasonable degree of certainty:

• The bold letters in the top banner change on refresh.
• The grid below is filled differently on refresh.
• Within any given square of the grid, the same text should always appear (conditioned on any text appearing at all inside that square).
• Some squares appear to be filled more often than others. In particular, one square seems to be always filled in with a label of FoF #I.
• The bolded letters seem to be replacing numbers.

From here, progress comes in the form of:

• Finding most or all of the FoFs and notating that data.
• Figuring out the letter to number correspondence.
• Understanding the rule(s) that determine which squares are filled.
• Understanding that the FoFs are either true or false (“fact or fiction”) and that it needs to be determined which is which.

All the data of what appears in each square is given below:

For understanding the letter to number correspondence, only the letters ABCILNORSY appear in bold. The puzzle title and banner text indicate that the number in the banner is describing some sort of view count (in fact, of the whole hunt website), and therefore it should only be increasing.

From this, we can find that each of these 10 letters are replacing a distinct (base-10) digit, and the correspondence, written from 0-9, is BINARYCOLS. (It is also possible to arrive at this conclusion by some sort of combination of guessing that the grid's FoF numbers are from 1 through 35, looking at some divisibility patterns in which FoFs appear with which, and guessing the string.)

The rule that governs which FoFs appear is: the FoF numbered [k] will appear if and only if [k] is a factor of the number that appears in the banner. This is also clued by the name “Fact or” Fiction.

With all of this in mind, it becomes possible to substitute all the bolded letters with numbers in the FoFs, and evaluate their truth values. Doing this yields

Most of these are straightforward once you know the numbering system and the factor rule, but some perhaps more complicated ones are elaborated on further in the appendix.

As indicated by the number to letter mapping of BINARYCOLS, we take the truth value of these FoFs in each column and read down, using false as 0 and true as 1, to get seven 5-bit values. Converting these values gives 1, 2, 19, 9, 14, 20, 8, which with the standard A-Z as 1-26 mapping reads as ABSINTH, a (variant spelling of a) type of (alcoholic) spirit.

If the solver calls this answer in, they are urged to “Keep going!”. The final step is to draw inspiration from the banner at the top of the page and to suppose that you were the ABSINTH visitor to the website, or the 30912th visitor. We find that 30912 is equal to 2^6 * 3 * 7 * 23, and so it has a lot of factors from 1 through 35. These factors are 1, 2, 3, 4, 6, 7, 8, 12, 14, 16, 21, 23, 24, 28, and 32. If we were to see the page as that visitor, only the cells highlighted below would be filled in:

Taking these cells that are filled in as 1s and the blank cells as 0s and again reading values as columns of binary gives us the final answer of GORDITA, a Mexican dish made with a stuffed tortilla.

Appendix

In this section, we go over some of the less straightforward FoFs.

FoF #19: There exist two not necessarily distinct prime numbers whose sum is [96345].FoF #29: There exist two not necessarily distinct prime numbers whose sum is [439638].

For the first of these, we note that two have two integers sum to an odd number, we need one of them to be even and one of them to be odd. Since the only even prime is 2, it suffices to check if 96345 - 2 = 96343 is prime or not. Since 96343 = 13 * 7411, it is not prime, and this FoF is false.

For the second of these, this FoF is a reference to Goldbach's conjecture. Though not known to be true, it has been verified to values up to 4 billion billion, so we can safely conclude that this FoF is true.

FoF #1: If this website is viewed [325] times per second starting from now, someone will be able to view a version of this page with all FoFs filled in within [165] years.

The first number which would cause all FoFs to be filled in is the least common multiple of all numbers from 1 through 35, or lcm(1, 2, …, 35) = 32 * 27 * 25 * 7 * 11 * 13 * 17 * 19 * 23 * 29 * 31 = 144403552893600 ≈ 1.44 * 10^14.

Using an estimate of 365.25 * 24 * 3600 seconds per year, and multiplying that by 325 visits per second and 165 years, we would have 1692276300000 ≈ 1.6 * 10^13 clicks, so that visit number would not yet have been reached. (Author’s note: should the counter ever reach 142711276593600, we will need to issue an errata.)

FoF #5: On average, it takes less than [32] views of this page to see a grid with only one FoF filled in.

For a page's number to cause only one FoF to be filled in, the number must be relatively prime to every prime below 35. Assuming a uniform distribution** modulo lcm(1, 2, …, 35), the probability for each prime can be calculated independently, and then all multiplied together. For any prime p, the counter has a (p-1)/p chance of being relatively prime to that number. Multiplying for each p < 35, we arrive at a probability of 1/2 * 2/3 * 4/5 * 6/7 * 10/11 * 12/13 * 16/17 * 18/19 * 22/23 * 28/29 * 30/31 ≈ 0.15285 that any given number will satisfy this property. It takes in expectation an average of 1/0.15285 ≈ 6.54 tries to get such a number.

FoF #11: On average, it takes less than [27] views of this page to see each FoF in the grid filled in at least one time.FoF #10: On average, it takes less than [218] views of this page to see each FoF in the grid filled in at least one time.

Again assuming a uniform distribution** modulo lcm(1, 2, …, 35), we can find by simulation that it takes around 35 trials to get 30 distinct FoFs, 53 to get 33 distinct FoFs, and around 93 to see all 35 FoFs, meaning the former of these FoFs is false and the latter is true.

We can also prove this average lies between the bounds with some bounding arguments. For the lower bound, since it takes an average of 35 trials to see FoF #35, it must take more than 27 trials to see everything.

For a handwavy upper bound, we first only consider the FoFs from 18 through 35, as if all of these are seen, then so must all the FoFs from 1 through 17. Note that for any FoF k, the probability that FoF hasn’t been seen within n trials is (1-1/k)^n. Choosing n=50, and summing over k from 18 through 35 we have an expected value of 2.627 unseen FoFs after 50 trials. (Wolfram Alpha) After these 50 trials, it should definitely take less than an average of 35+35+35 to find these remaining unseen FoFs, which will leave us well below the 218 upper bound.

**The author recognizes that the puzzle does not exactly make clear what the probability distribution on the counter ought to be, but thinks that anyone pedantic enough to notice that issue ought to be able to deduce the answer anyway. :)

Author’s Notes

I originally designed this puzzle for the intro round. In this initial version, the counter would count views only to that page (not the entire website), and it would be unlocked at the beginning of the hunt. For all teams. This would invite hunters to all DDoS our website at the same time right at the beginning of hunt. I was urged by several other members of our development team not to do this. As you can see, this did not end up happening, which was probably for the best.

This puzzle originally didn’t have the final ABSINTH step–that was something I happened to stumble upon while trying to make the extraction better than just reading off binary in the columns. It’s a bit of a minor miracle that it ends up working out.

Many many thanks to Max and Benji for working with me to find an answer that fits the constraints of the CTO meta and also the final extraction step of this puzzle. They ended up reworking a mechanic of the meta and replacing all of its answers, and I wound up with first dibs on an answer that worked for me. :)