Find the number of the element of this set. This is the question that someone give me. 
Let the set, $S = \{ n | p \leq  3^{-n}q \lt 243p\}$ for a integer, $n$ and the positive real numbers $p,q$
Find the number of the element the set,$ S$

I thought the answer is countable infinity. But the someone who gavemethis questio  said the answer is 4. 
It looks like the above question weird I thought. 
To conclude the answer is 4, Do we need more condition, don't we?
What do you think? 
Any opinion would be help. Thanks.
 A: The answer is that $|S|=5$ in all cases. Note that unlike what's been written in the comments, $3^4=81; 3^5=243$.
Contrary to often used naming convention, as stated $n$ can be any integer, not just a natural number.
The left part of the inequality for $n$ can be equivalently transformed as follows, considering that $p$ and $q$ are assumed to be positive:
$$
\begin{eqnarray} 
p                  & \le & 3^{-n}q  &\Longleftrightarrow& \\
\frac{p}q          & \le & 3^{-n}   &\Longleftrightarrow& \\
\log_3\frac{p}q    & \le & -n       &\Longleftrightarrow& \\
n                  & \le & -\log_3\frac{p}q.     \\
\end{eqnarray}
$$
The same can be done for the right part of the inequality (I leave out the equivalent steps as done above):
$$
\begin{eqnarray} 
3^{-n}q    & < &  243p               &\Longleftrightarrow& \\
n          & > & -\log_3\frac{243p}q &\Longleftrightarrow& \\
n          & > & -\log_3\frac{p}q - \log_3 243 &\Longleftrightarrow& \\
n          & > & -\log_3\frac{p}q - 5. \\
\end{eqnarray}
$$
So we find that the condition given in the definition of $S$ is equivalent to
$$ -\log_3\frac{p}q - 5 < n \le -\log_3\frac{p}q$$
No matter what exact value $\log_3\frac{p}q$ has, there are always exactly 5 integer values of $n$ that fullfill the above condition:
$$\lfloor-\log_3\frac{p}q\rfloor -k$$
for $k=0,1,2,3,4$.
