2013 AMC 12B Problem 18

Question

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

or

4

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

1

or

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?

Barbara will win with

2013

coins, and Jenna will win with

2014

coins.

Jenna will win with

2013

coins, and whoever goes first will win with

2014

coins.

Barbara will win with

2013

coins, and whoever goes second will win with

2014

coins.

Jenna will win with

2013

coins, and Barbara will win with

2014

coins.

Whoever goes first will win with

2013

coins, and whoever goes second will win with

2014

coins.

Answer: B

Concepts:combinatorial game invariant modular arithmetic

Difficulty rating: 2070

Small hint:

Track the number of coins modulo

5

Big hint:

Jenna can always restore a multiple of

5

after her move, answering Barbara’s

2

with

3

and her

4

with

1;

decide who can maintain the invariant for each starting count

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.

2013 AMC 12B Problem 18

Answer: B

Difficulty rating: 2070

Small hint:

Big hint:

Solution:

Problem 18 in other years