AMC10 2012 B

AMC10 2012 B · Q20

AMC10 2012 B · Q20. It mainly tests Linear equations, Games (basic).

Bernardo and Silvia play the following game. An integer between 0 and 999, inclusive, is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds 50 to it and passes the result to Bernardo. The winner is the last person who produces a number less than 1000. Let N be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of N ?

贝尔纳多和西尔维娅玩以下游戏。选一个0到999（含）的整数给贝尔纳多。贝尔纳多收到数字时加倍后传给西尔维娅。西尔维娅收到数字时加50后传给贝尔纳多。产生小于1000的数字的最后一人获胜。令N为导致贝尔纳多获胜的最小初始数。N各位数字之和是多少？

(A) 7 7

(B) 8 8

(D) 10 10

(E) 11 11

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): The smallest initial number for which Bernardo wins after one round is the smallest integer solution of $2n + 50 \ge 1000$, which is 475. The smallest initial number for which he wins after two rounds is the smallest integer solution of $2n + 50 \ge 475$, which is 213. Similarly, the smallest initial numbers for which he wins after three and four rounds are 82 and 16, respectively. There is no initial number for which Bernardo wins after more than four rounds. Thus $N = 16$, and the sum of the digits of $N$ is 7.

答案（A）：伯纳多在一轮后获胜所需的最小初始数，是不等式 $2n + 50 \ge 1000$ 的最小整数解，即 475。两轮后获胜所需的最小初始数，是不等式 $2n + 50 \ge 475$ 的最小整数解，即 213。同理，三轮和四轮后获胜所需的最小初始数分别是 82 和 16。不存在使伯纳多在超过四轮后获胜的初始数。因此 $N = 16$，且 $N$ 的各位数字之和为 7。

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