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This is a question from the 2012 Tournament of Towns Spring Senior O2: “The cells of a $1\times 2n$ board are labelled $1,2,...,,n,- n,...,- 2,-1$ from left to right. A marker is placed on an arbitrary cell. If the label of the cell is positive, the marker moves to the right a number of cells equal to the value of the label. If the label is negative, the marker moves to the left a number of cells equal to the absolute value of the label. Prove that if the marker can always visit all cells of the board, then $2n + 1$ is prime.”

This is what I know so far — The cells on the board always lead to another designated cell, i.e. $1$ always leads to $2$ and $2$ always leads to $4$. Since we want to show that the marker can always visit all cells of the board, we can show that it can visit all cells of the board starting from any number (since the moves form a closed loop). Then I discovered that for some numbers like $7$, we can choose $5$ and it forms a closed loop in one move. ($5$ to $-5$, $-5$ to $5$). I don’t know how to continue from here though. It would be very nice if I could receive a hint.

Thank you

Protein Rocket
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    I believe you quoted the wrong problem. I think you want $2n+1$ is prime. See the source – Calvin Lin Jan 01 '25 at 20:46
  • Re your reasoning, we have a walk of length $2n-1$, but that doesn't mean that the final square closes back onto the first square. You need to show that each square leads to a unique square, which might be what you meant by "lead to another designated cell"? – Calvin Lin Jan 01 '25 at 20:52
  • Each step corresponds to doubling the value $!\bmod 2n+1,,$ so if the doubling map visits all cells then $2$ has order $2n$ so $2n+1$ is prime, by the dupe. – Bill Dubuque Jan 01 '25 at 20:53
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    Update: @Calvin linked the official solution. The idea is that if the modulus $,m = 2n!+!1 = dk,$ is composite with nontrivial factor $,d>1,,$ then if we start at a cell $,dj,$ then the factor $,d,$ remains when doubling (by $,2dj\bmod dk = d:!(2j\bmod k),,$ by mod distributive law), so we can visit (at most) the $,k = m/d,$ cells that are multiples of $,d.\ \ $ – Bill Dubuque Jan 01 '25 at 21:38
  • Thanks for the hints, should I delete this post or leave it as it is? – Protein Rocket Jan 02 '25 at 10:15

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