AMC8 1990

AMC8 1990 · Q21

AMC8 1990 · Q21. It mainly tests Sequences & recursion (algebra).

A list of 8 numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three are shown: \[ ? , \underline{\hspace{1cm}} , \underline{\hspace{1cm}} , \underline{\hspace{1cm}} , \underline{\hspace{1cm}} , \underline{\hspace{1cm}} , 16 , 64 , 1024 \] what is the first number?

一个由8个数组成的列表，从给定的两个数开始形成。列表中的每个新数是前两个数的乘积。如果最后三个数已知： \[? , \underline{\hspace{1cm}} , \underline{\hspace{1cm}} , \underline{\hspace{1cm}} , \underline{\hspace{1cm}} , \underline{\hspace{1cm}} , 16 , 64 , 1024\] 求第一个数。

(A) \frac{1}{64} \frac{1}{64}

(B) \frac{1}{4} \frac{1}{4}

(D) 2 2

(E) 4 4

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Working backward from $1024$, divide each number (e.g. $1024$) by the preceding number (e.g. $64$) to get the previous number (e.g. $16$) in the list. Thus $64 \div 16 = 4$, $4 \div 4 = 1$, and so on: $\frac{1}{4},\ 4,\ 1,\ 4,\ 4,\ 16,\ 64,\ 1024$

答案（B）：从 $1024$ 开始向前推，用当前数（例如 $1024$）除以前一个数（例如 $64$），即可得到再前一个数（例如 $16$）。因此 $64 \div 16 = 4$，$4 \div 4 = 1$，依此类推： $\frac{1}{4},\ 4,\ 1,\ 4,\ 4,\ 16,\ 64,\ 1024$

Topics

AL.012 Sequences & recursion (algebra)