(Post USAMO update)

Predicted distribution: 22 = 7/7/1 7/0/0

Solution sketches/progress:

p1: I let p be the first prime larger than n, and out of p+1, p+2, p+3, p+4, and p+5, there is a multiple of 3 and a multiple of 4 that are consecutive. Furthermore, for n>=11 i showed using Bertrand's that any multiples of 3 or 4 at this "size level" must divide n!, since it is split to either 3 times something at most n or 4 times something at most n, and 9 and 16 also divide n!.

p2: I rephrased the problem like this: there is a vertex corresponding to each subset of {1,2,3...100}, such that a vertex of size k has k "colors" that it cycles between (essentially a mod k tracker). and in each move, you "tap" a vertex, and all of its subsets change color.

The tapping is analogous to introducing a new element and throwing it in the sets corresponding to the tapped vertex.

I showed that tapping a vertex of size k k times is the same as tapping each of its k subsets of size k-1 once each, which is the main idea. From here, it is not too hard to get the bound and construction as you can keep sending taps down until each of the 50-size subsets are forced a tap and thus 50 taps each.

p3: I claimed the correct answer and provided a construction (triangulate at one vertex, color mod m) as well as proved that the construction works.

I used the 1/2 ab sin C area rule, to reduce it to a sum of sin kpi/n (k+1)pi/n where k is a fixed residue mod m. By product to sum, this splits into a term that does not depend on what the residue is mod m and another term that is just always zero since it sums equally spaced points around the unit circle.

p4: I got m <= n+1, and the construction I had was n red 0 blue, n-1 red 1 blue, n-2 red 2 blue, and so on until n-m+1 red m-1 blue. I showed this works by putting them in a grid of m rows and n columns and noting that shifting the starting index forward by 1 is equivalent to cyclic shifting one of the columns of the grid up, and we can use this to show that the last row gains a red while some other row loses one, causing the last row's score to swap with the row that lost one, so they are all distinct.

p5: Tried inverting at B and stated the inverted problem, but did not end up solving.

p6: Claimed wrong answer lol.

Important note: due to concerns about Manifold contributing to illegal discussion, as there have been instances in the past where bets were made on markets about a competition during the anti-discussion period, this market will close (but not resolve) temporarily before USAMO. Once discussion is actually allowed, I will re-open the market and allow for bets again.

Resolves N/A if the award system for the USAMO changes (which may happen, especially considering that it has changed a lot recently)

Clarification: HM is between 14 and bronze cutoff