AMC12 2012 B

AMC12 2012 B · Q14

AMC12 2012 B · Q14. It mainly tests Rates (speed), 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 $N$?

Bernardo 和 Silvia 玩如下游戏。从 $0$ 到 $999$（含）之间选取一个整数并给 Bernardo。每当 Bernardo 收到一个数，他将其加倍并把结果传给 Silvia。每当 Silvia 收到一个数，她将其加 $50$ 并把结果传给 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？

(A) 7 7

(B) 8 8

(D) 10 10

(E) 11 11

Answer

Correct choice: (A)

正确答案：（A）

Solution

The last number that Bernardo says has to be between 950 and 999. Note that $1\rightarrow 2\rightarrow 52\rightarrow 104\rightarrow 154\rightarrow 308\rightarrow 358\rightarrow 716\rightarrow 766$ contains 4 doubling actions. Thus, we have $x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \rightarrow 8x+300 \rightarrow 8x+350 \rightarrow 16x+700$. Thus, $950<16x+700<1000$. Then, $16x>250 \implies x \geq 16$. Because we are looking for the smallest integer $x$, $x=16$. Our answer is $1+6=\boxed{7}$, which is A. Work backwards. The last number Bernardo produces must be in the range $[950,999]$. That means that before this, Silvia must produce a number in the range $[475,499]$. Before this, Bernardo must produce a number in the range $[425,449]$. Before this, Silvia must produce a number in the range $[213,224]$. Before this, Bernardo must produce a number in the range $[163,174]$. Before this, Silvia must produce a number in the range $[82,87]$. Before this, Bernardo must produce a number in the range $[32,37]$. Before this, Silvia must produce a number in the range $[16,18]$. Silvia could not have added 50 to any number before this to obtain a number in the range $[16,18]$, hence the minimum $N$ is 16 with the sum of digits being $\boxed{\textbf{(A)}\ 7}$.

Bernardo 最后说出的数必须在 950 到 999 之间。注意到 $1\rightarrow 2\rightarrow 52\rightarrow 104\rightarrow 154\rightarrow 308\rightarrow 358\rightarrow 716\rightarrow 766$ 包含 4 次加倍操作。因此有 $x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \rightarrow 8x+300 \rightarrow 8x+350 \rightarrow 16x+700$。因此，$950<16x+700<1000$。于是 $16x>250 \implies x \geq 16$。由于我们要找最小整数 $x$，所以 $x=16$。答案为 $1+6=\boxed{7}$，对应 A。倒推：Bernardo 产生的最后一个数必须在区间 $[950,999]$。这意味着在此之前，Silvia 必须产生一个在区间 $[475,499]$ 的数。在此之前，Bernardo 必须产生一个在区间 $[425,449]$ 的数。在此之前，Silvia 必须产生一个在区间 $[213,224]$ 的数。在此之前，Bernardo 必须产生一个在区间 $[163,174]$ 的数。在此之前，Silvia 必须产生一个在区间 $[82,87]$ 的数。在此之前，Bernardo 必须产生一个在区间 $[32,37]$ 的数。在此之前，Silvia 必须产生一个在区间 $[16,18]$ 的数。Silvia 不可能通过在此之前对某个数加 50 得到区间 $[16,18]$ 内的数，因此最小的 $N$ 为 16，其各位数字之和为 $\boxed{\textbf{(A)}\ 7}$。

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