Question

Bernardo 和 Silvia 玩下面的游戏。先选一个 $0$ 到 $999$ 之间的整数给 Bernardo。每当 Bernardo 收到一个数，他就把它加倍并传给 Silvia。每当 Silvia 收到一个数，她就加上 $50$ 并传给 Bernardo。最后一个产生小于 $1000$ 的数的人获胜。

令 $N$ 为能使 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：递推不等式数字

难度评级：1930

解答：

设初始数为 $x$ 。传给 Silvia 后是 $50$ ，Silvia 再传回 Bernardo 后是。继续迭代，某次传给 Silvia 的数为，Silvia 产生的数为，再往后就会超过 $1000$ 。为使 Bernardo 是最后一个产生小于 $1000$ 的数的人，传给 Silvia 的数必须小于，而 Silvia 产生的数必须大于它： $\begin{gathered} 2x,\quad4x+100,\quad8x+300,\\ 16x+700,\quad32x+1500. \end{gathered}$

因此 $x$ ，所以这个整数满足 $475,213,82$ ，最小的 $N=16$ 。于是 $16$ 的各位数字之和为 $1+6=7$ 。 $1500$ 。

所以正确答案是 A。

If the initial number is $x,$ Bernardo's successive outputs are $\begin{gathered} 2x,\quad4x+100,\quad8x+300,\\ 16x+700,\quad32x+1500. \end{gathered}$ Silvia's output after each of these is $50$ larger. Bernardo wins on a given turn exactly when his output is below $1000$ but the following output from Silvia is at least $1000.$

For the first four Bernardo turns, the smallest integer $x$ satisfying those two inequalities is, respectively, $475,213,82,$ and $16.$ A fifth Bernardo output is already at least $1500.$ Thus the smallest winning initial number is $N=16,$ whose digit sum is $1+6=7.$

Thus, the correct answer is A .

2012 AMC 10B 第 20 题

答案：A

难度评级：1930

解答：

其他年份的第 20 题