Question

Bernardo 和 Silvia 玩下面的游戏。选取一个从 $0$ 到 $999$ （含两端）的整数交给 Bernardo。每当 Bernardo 收到一个数时，他把它加倍并把结果交给 Silvia。每当 Silvia 收到一个数时，她给它加上 $50$ 并把结果交给 Bernardo。最后一个产生小于 $1000$ 的数的人获胜。设 $N$ 为使 Bernardo 获胜的最小初始数。 $N$ 的各位数字之和是多少？

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 $N?$

$7$

$8$

$9$

$10$

$11$

答案：A

知识点：逆推法不等式

难度评级：1730

解答：

当 Bernardo 在一轮后给出的加倍数 $2n+50\ge1000$ 且之前的数都小于 $1000$ 时，他获胜。满足 $2n+50\ge1000$ 的最小 $n$ 是 $475$ 。

倒推，经过二、三、四轮后导致胜利的最小初始值分别是满足 $2n+50\ge475$ ， $\ge213$ ，和 $\ge82$ 的最小整数，即 $213$ ， $82$ ，和 $16$ 。不会有超过四轮后才获胜的初始值。

所以 $N=16$ ，各位数字之和为 $1+6=7$ 。

因此正确答案是 A。

Bernardo wins after a round when his doubled number $2n+50\ge1000$ but the previous numbers stayed below $1000.$ The smallest $n$ with $2n+50\ge1000$ is $475.$

Working backwards, the smallest starting values that lead to a win after two, three, and four rounds are the smallest integers with $2n+50\ge475,$ $\ge213,$ and $\ge82,$ namely $213,$ $82,$ and $16.$ No start wins after more than four rounds.

So $N=16,$ and the sum of its digits is $1+6=7.$

Thus, the correct answer is A.

2012 AMC 12B 第 14 题

答案：A

难度评级：1730

解答：

其他年份的第 14 题