AMC12 2016 B

AMC12 2016 B · Q25

AMC12 2016 B · Q25. It mainly tests Exponents & radicals, Sequences & recursion (algebra).

The sequence $(a_n)$ is defined recursively by $a_0=1$, $a_1=\sqrt[19]{2}$, and $a_n=a_{n-1}a_{n-2}^2$ for $n\ge 2$. What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?

数列$(a_n)$按递推方式定义：$a_0=1$，$a_1=\sqrt[19]{2}$，并且当$n\ge 2$时，$a_n=a_{n-1}a_{n-2}^2$。求使得乘积$a_1a_2\cdots a_k$为整数的最小正整数$k$。

(A) 17 17

(B) 18 18

(D) 20 20

(E) 21 21

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Express each term of the sequence $(a_n)$ as $2^{b_n/19}$. (Equivalently, let $b_n$ be the logarithm of $a_n$ to the base $\sqrt[19]{2}$.) The recursive definition of the sequence $(a_n)$ translates into $b_0=0$, $b_1=1$, and $b_n=b_{n-1}+2b_{n-2}$ for $n\ge 2$. Then the product $a_1a_2\cdots a_k$ is an integer if and only if $\sum_{i=1}^k b_i$ is divisible by $19$. Let $c_n=b_n\ \bmod\ 19$. It follows that $a_1a_2\cdots a_k$ is an integer if and only if $p_k=\sum_{i=1}^k c_i$ is divisible by $19$. Let $q_k=p_k\ \bmod\ 19$. Because the largest answer choice is $21$, it suffices to compute $c_k$ and $q_k$ successively for $k$ from $1$ up to at most $21$, until $q_k$ first equals $0$. The modular computations are straightforward from the definitions. \[ \begin{array}{c|ccccccccccccccccc} k & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17\\ \hline c_k & 1&1&3&5&11&2&5&9&0&18&18&16&14&8&17&14&10\\ q_k & 1&2&5&10&2&4&9&18&18&17&16&13&8&16&14&9&0 \end{array} \] Thus the requested answer is $17$.

答案（A）：将数列$(a_n)$的每一项表示为$2^{b_n/19}$。（等价地，令$b_n$为$a_n$以$\sqrt[19]{2}$为底的对数。）数列$(a_n)$的递推定义可转化为$b_0=0$、$b_1=1$，并且当$n\ge 2$时有$b_n=b_{n-1}+2b_{n-2}$。那么，乘积$a_1a_2\cdots a_k$为整数当且仅当$\sum_{i=1}^k b_i$能被$19$整除。令$c_n=b_n\ \bmod\ 19$。因此，$a_1a_2\cdots a_k$为整数当且仅当$p_k=\sum_{i=1}^k c_i$能被$19$整除。令$q_k=p_k\ \bmod\ 19$。由于备选答案中最大为$21$，只需对$k$从$1$起依次计算$c_k$与$q_k$，最多到$21$，直到首次出现$q_k=0$为止。按定义进行模运算计算即可。 \[ \begin{array}{c|ccccccccccccccccc} k & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17\\ \hline c_k & 1&1&3&5&11&2&5&9&0&18&18&16&14&8&17&14&10\\ q_k & 1&2&5&10&2&4&9&18&18&17&16&13&8&16&14&9&0 \end{array} \] 因此所求答案为$17$。

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