2013 AMC 10B Problem 17

Question

Alex has

75

red tokens and

75

blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?

62

82

83

102

103

Answer: E

Concepts:Diophantine Equation invariant process simulation

Difficulty rating: 1970

Small hint:

Let

m

and

n

be the numbers of the two exchange types

Big hint:

No more moves means fewer than

2

red and fewer than

3

blue tokens

Solution:

Suppose Alex makes

m

exchanges at the red-token booth and

n

exchanges at the blue-token booth.

He then has

75-2m+n

red tokens and

75+m-3n

blue tokens. At the end he must have fewer than

2

red tokens and fewer than

3

blue tokens.

Solving these terminal possibilities gives only two candidate final token counts:

(1,2)

, which comes from

(m,n)=(59,44)

, or

(0,0)

, which comes from

(m,n)=(60,45)

.

The final count

(0,0)

is impossible, because the last exchange would always create either one blue token or one red token.

The final count

(1,2)

is attainable. Starting from

(75,75)

red and blue tokens, make

25

blue-booth exchanges, then

50

red-booth exchanges, then

16

blue-booth exchanges, then

8

red-booth exchanges, then

3

blue-booth exchanges, and finally

1

red-booth exchange. The red-blue counts become

\begin{aligned} &(100,0),(0,50),(16,2),\\ &(0,10),(3,1),(1,2). \end{aligned}

Therefore Alex ends with

59+44=103

silver tokens, and the correct answer is E .