2013 AMC 12B Problem 18

Problem 18 of 25IntermediateNumber Theory Combinatorics Problem-Solving Techniques

Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove

2

4

coins, unless only one coin remains, in which case she loses her turn. When it is Jenna’s turn, she must remove

1

3

coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with

2013

coins and when the game starts with

2014

coins?

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Solution

Work modulo

5.

With

2013 \equiv 3

coins, Jenna wins either way: going first she takes

3

to leave a multiple of

5,

then answers Barbara’s

2

with

3

and

4

with

1

to keep multiples of

5,

eventually taking the last coin; going second she keeps the count

\equiv 3 \pmod 5

until Barbara is stuck at

3

coins, must remove

2,

and leaves Jenna the last coin. With

2014 \equiv 4

coins, whoever goes first wins: Jenna first reduces to the

2013

case, while Barbara first takes

4

and then keeps multiples of

5.

This is choice B. Thus, the correct answer is B.

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Tagged: combinatorial game · invariant · modular arithmetic

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