Ponder This Challenge - April 2021 - Decremental wheel of choice

A wheel of choice is marked with the numbers 1,2,...,n, and a prize is associated with each number.

A player faced with the wheel chooses some number q and starts spinning the wheel k times. On each spin, the wheel moves forward q steps in a clockwise direction and the number / prize reached is eliminated from the wheel (i.e., the player does not get it). After k such spins, all the prizes remaining on the wheel are gifted to the player.
At the beginning, the wheel's arrow points between 1 and n. If the wheel reaches the number t, t is removed from the wheel, the size of the remaining numbers are readjusted and evenly spaced, and the arrow is moved to the point in between t's left and right neighbors. Each step moves the wheel a little less than one number forward (so the wheel comes to rest on a number and not between two). So, if the wheel is right before 1 and performs 3 steps, it ends up on 3.

An example of the wheel for n=8:

And after three spins with q=5:

For a given n, we call a set of k numbers unwinnable if no matter what q the player chooses, he will not be able to win exactly this set after n-k spins. The game show wants to know the unwinnable sets of a given n to avoid handing out the best prizes.

For example, there are no unwinnable sets for n smaller than 9. For n=9, there are three unwinnable sets, all of size k=5:

Your goal: Find an n such that there is a set of unwinnable numbers for seven steps (i.e., the set is of size n-7). In your answer, supply the number n and the elements of the unwinnable set.

A bonus '*' for finding an unwinnable set which takes more than seven steps.

Solution

• Click here to view the solution

  For n = 20, there are two sets with n-7 = 13 elements that are unwinnable:

  (1, 2, 3, 4, 5, 6, 7, 8, 11, 14, 15, 16, 17)
  (4, 5, 6, 7, 10, 13, 14, 15, 16, 17, 18, 19, 20)
  

  For n = 21, there are many sets with n-8 = 13 elements that are unwinnable. For example:

  (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15)
  

Solvers

• *Lorenz Reichel (1/4/2021 10:57 PM IDT)
• *Guillaume Escamocher (1/4/2021 11:27 PM IDT)
• *Uoti Urpala (1/4/2021 11:55 PM IDT)
• *Victor Chang (2/4/2021 3:19 AM IDT)
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• *Phil Proudman (2/4/2021 6:36 PM IDT)
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• *Alper Halbutogullari (2/4/2021 10:03 PM IDT)
• *Dominik Reichl (2/4/2021 11:37 PM IDT)
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