AMC8 2020

AMC8 2020 · Q22

AMC8 2020 · Q22. It mainly tests Manipulating equations, Patterns & sequences (misc).

When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.

当一个正整数 $N$ 输入机器时，输出是一个根据下面规则计算的数字。

(A) 73 73

(B) 74 74

(D) 82 82

(E) 83 83

Answer

Correct choice: (E)

正确答案：（E）

Solution

We start with final output of $1$ and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage: \[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}\] where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$), but $16$ could come from $32$ or $5$ (as $\frac{32}{2} = 3 \cdot 5 + 1 = 16$, and $32$ is even while $5$ is odd). By construction, last set in this sequence contains all the numbers which will lead to number $1$ to end of the $6$-step process, and sum is $1+8+10+64=\boxed{\textbf{(E) }83}$.

我们从最终输出 $1$ 开始，向后推导，注意考虑所有可能导致特定输出的输入。这样在每个阶段产生以下可能性集合： \[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}\] 其中，例如，$2$ 必须来自 $4$（因为没有整数 $n$ 满足 $3n+1=2$），但 $16$ 可以来自 $32$ 或 $5$（因为 $\frac{32}{2} = 3 \cdot 5 + 1 = 16$，且 $32$ 是偶数而 $5$ 是奇数）。通过这种构造，该序列的最后一个集合包含所有经过 $6$ 步过程最终导致 $1$ 的数字，其和是 $1+8+10+64=\boxed{\textbf{(E) }83}$。

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