AMC12 2013 B

AMC12 2013 B · Q10

AMC12 2013 B · Q10. It mainly tests Systems of equations, Diophantine equations (integer solutions).

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?

Alex 有 75 个红代币和 75 个蓝代币。有一个摊位，Alex 可以交 2 个红代币换取 1 个银代币和 1 个蓝代币；另一个摊位，可以交 3 个蓝代币换取 1 个银代币和 1 个红代币。Alex 继续交换直到无法再交换。最终 Alex 有多少银代币？

(A) 62 62

(B) 82 82

(D) 102 102

(E) 103 103

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): After Alex makes $m$ exchanges at the first booth and $n$ exchanges at the second booth, Alex has $75-(2m-n)$ red tokens, $75-(3n-m)$ blue tokens, and $m+n$ silver tokens. No more exchanges are possible when he has fewer than $2$ red tokens and fewer than $3$ blue tokens. Therefore no more exchanges are possible if and only if $2m-n\ge 74$ and $3n-m\ge 73$. Equality can be achieved when $(m,n)=(59,44)$, and Alex will have $59+44=103$ silver tokens. Note that the following exchanges produce $103$ silver tokens: \[ \begin{array}{l|c|c|c} & \text{Red Tokens} & \text{Blue Tokens} & \text{Silver Tokens}\\ \hline \text{Exchange 75 blue tokens} & 100 & 0 & 25\\ \text{Exchange 100 red tokens} & 0 & 50 & 75\\ \text{Exchange 48 blue tokens} & 16 & 2 & 91\\ \text{Exchange 16 red tokens} & 0 & 10 & 99\\ \text{Exchange 9 blue tokens} & 3 & 1 & 102\\ \text{Exchange 2 red tokens} & 1 & 2 & 103 \end{array} \]

答案（E）：在 Alex 在第一个摊位交换 $m$ 次、在第二个摊位交换 $n$ 次之后，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。当他的红色代币少于 $2$ 个且蓝色代币少于 $3$ 个时，就无法再继续交换。因此，当且仅当 $2m-n\ge 74$ 且 $3n-m\ge 73$ 时，不能再进行交换。取等号可在 $(m,n)=(59,44)$ 时实现，此时 Alex 将有 $59+44=103$ 个银色代币。注意：下面这些交换过程会得到 $103$ 个银色代币： \[ \begin{array}{l|c|c|c} & \text{红色代币} & \text{蓝色代币} & \text{银色代币}\\ \hline \text{交换 75 个蓝色代币} & 100 & 0 & 25\\ \text{交换 100 个红色代币} & 0 & 50 & 75\\ \text{交换 48 个蓝色代币} & 16 & 2 & 91\\ \text{交换 16 个红色代币} & 0 & 10 & 99\\ \text{交换 9 个蓝色代币} & 3 & 1 & 102\\ \text{交换 2 个红色代币} & 1 & 2 & 103 \end{array} \]

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