Question

Alex 有 $75$ 个红色代币和 $75$ 个蓝色代币。有一个摊位可以用两个红色代币换一个银色代币和一个蓝色代币；另一个摊位可以用三个蓝色代币换一个银色代币和一个红色代币。Alex 持续交换，直到无法再交换为止。最后 Alex 会有多少个银色代币？

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?

$62$

$82$

$83$

$102$

$103$

答案：E

知识点：方程组不变量

难度评级：1550

解答：

经过 $m$ 次红色摊位交换和 $n$ 次蓝色摊位交换后，Alex 有 $75 - (2m - n)$ 个红色代币、 $75-(3n-m)$ 个蓝色代币，以及 $m+n$ 个银色代币。停止时，红色代币数为 $0$ 或 $1,$ 蓝色代币数为 $0,1,$ 或 $2.$ 对这六种情形求解代币方程，只得到 $(m,n)=(59,44),$ 终态为 $(1,2),$ 或 $(m,n)=(60,45),$ 终态为 $(0,0).$ 后者不可能到达：最后一次交换前的状态只能是 $(-1,3)$ 或 $(2,-1).$ 因此 Alex 最终有 $59+44=103$ 个银色代币。所以正确答案是 E。

After $m$ red-booth and $n$ blue-booth exchanges, Alex has $75 - (2m - n)$ red tokens, $75-(3n-m)$ blue tokens, and $m+n$ silver tokens. At termination the red count is $0$ or $1,$ and the blue count is $0,1,$ or $2.$ Solving the token equations over these six cases leaves only $(m,n)=(59,44),$ ending at $(1,2),$ or $(m,n)=(60,45),$ ending at $(0,0).$ The latter is unreachable: its final exchange would have to start at either $(-1,3)$ or $(2,-1).$ Hence Alex finishes with $59+44=103$ silver tokens. Thus, the correct answer is E.

2013 AMC 12B 第 10 题

答案：E

难度评级：1550

解答：

其他年份的第 10 题