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2020 AMC 10B Problem 16

2020 AMC 10B Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Games & Processes · difficulty 3 of 5 · about 3 min · position P16-20

Bela and Jenn play the following game on the closed interval $[0, n]$ of the real number line, where $n$ is a fixed integer greater than $4$ . They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval $[0, n]$ . 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?

A)
$\text{ Bela will always win.}$
B)
$\text{ Jenn will always win.}$
C)
$\text{ Bela will win if and only if }n \text{ is odd.}$
D)
$\text{ Jenn will win if and only if }n \text{ is odd.}$
E)
$\text { Jenn will win if and only if } n>8.$

Answer

Key insight

Bela takes the midpoint n/2 first, then mirrors every Jenn move across it; the mirror move is always legal, so Jenn runs out first.

Solution

Bela starts by choosing the midpoint $\tfrac n2$ . From then on, whenever Jenn chooses a number $x$ , Bela answers with its mirror image $n - x$ .

Why this is always legal: the set of chosen numbers is symmetric about $\tfrac n2$ after each of Bela's turns. If Jenn's $x$ is more than $1$ from every chosen number, then $n - x$ is more than $1$ from every chosen number as well (distances are preserved by reflection). And $n - x \ne x$ , since $x = \tfrac n2$ is already taken and any legal $x$ is more than $1$ away from it; also $n - x$ is not within $1$ of $x$ , because $|x - (n-x)| = 2|x - \tfrac n2| > 2$ .

So Bela always has a move after Jenn does. The game ends (choices are spaced more than $1$ apart on a bounded interval, so only finitely many fit), and the player who first cannot move is necessarily Jenn.

The answer is $\boxed{\textbf{(A)}\ \text{Bela will always win.}}$

Why this works

Symmetry strategies win games where the board has a center: the first player occupies the center, then copies the opponent's move reflected through it. The proof has two parts, and both matter: the mirrored move is legal, and it never coincides with (or conflicts with) the move just made. The parity of $n$ is a distraction; the answer choices mentioning odd/even are bait.

The trap

Trying to analyze the count of remaining moves as a function of n (odd versus even) instead of looking for a strategy-stealing symmetry.

Common mistakes

Trying to analyze the count of remaining moves as a function of n (odd versus even) instead of looking for a strategy-stealing symmetry.
Believing Jenn can also mirror; she cannot, because the midpoint is already gone and the first player owns the symmetry.

Techniques

Exploit symmetry to reduce work or pair up objects