2020 AMC 10B Problems/Problem 16

The following problem is from both the 2020 AMC 10B #16 and 2020 AMC 12B #14, so both problems redirect to this page.


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?

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


Notice that to use the optimal strategy to win the game, Bela must select the middle number in the range $[0, n]$ and then mirror whatever number Jenn selects. Therefore, if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so the answer is $\boxed{\textbf{(A)} \text{ Bela will always win}}$.

Solution 2 (Guessing)

First of all, realize that the value of $n$ should have no effect on the strategy at all. This is because they can choose real numbers, not integers, so even if $n$ is odd, for example, they can still go halfway. Similarly, there is no reason the strategy would change when $n>8$.

So we are left with (A) and (B). From here it is best to try out random numbers and try to find the strategy that will let Bela win, but if you can't find it, realize that it is more likely the answer is $\boxed{\textbf{(A)} \text{ Bela will always win}}$ since Bela has the first move and thus has more control.

Video Solution

https://youtu.be/3BvJeZU3T-M (for AMC 10) https://youtu.be/0xgTR3UEqbQ (for AMC 12)


See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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