2020 AMC 12B Problem 14

Below is the professionally curated solution for Problem 14 of the 2020 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12B solutions, or check the answer key.

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Concepts:combinatorial gamesymmetry

Difficulty rating: 1500

14.

Bela and Jenn play the following game on the closed interval [0,n][0, n] of the real number line, where nn is a fixed integer greater than 4.4. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval [0,n].[0, n]. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?

Bela will always win.

Jenn will always win.

Bela will win if and only if nn is odd.

Jenn will win if and only if nn is odd.

Jenn will win if and only if n>8.n \gt 8.

Solution:

Bela first plays the midpoint n2.\tfrac{n}{2}. This choice makes the configuration symmetric about the center of the interval.

Thereafter, whenever Jenn picks a number x,x, Bela responds with its mirror image nx.n - x. Since the position was symmetric before Jenn moved and her move is legal, its reflection is also legal and distinct. Thus Bela always has a move whenever Jenn does, so Jenn is the first to be stuck. Bela always wins.

Thus, the correct answer is A.

Problem 14 in Other Years