# 1990 AJHSME Problems/Problem 21

## Problem

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: $$\text{\underline{\hspace{3 mm}?\hspace{3 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm}\underline{\hspace{2 mm}16\hspace{2 mm}}\hspace{1 mm},\hspace{1 mm}\underline{\hspace{2 mm}64\hspace{2 mm}}\hspace{1 mm},\hspace{1 mm}\underline{\hspace{1 mm}1024\hspace{1 mm}}}$$ $\text{(A)}\ \frac{1}{64} \qquad \text{(B)}\ \frac{1}{4} \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 2 \qquad \text{(E)}\ 4$

## Solution

We just use the definition to find the first number is $1/4 \rightarrow \boxed{\text{B}}$.

## Solution 2

If we do $64/16$ we get $4$, then we do $16/4$ giving $4$ again, then $4/4$ is $1$ and $4/1$ is $4$ and finally $1/4$ gives $1/4$ as the answer. Which is $\boxed{\text{B}}$