Problem source

Let $a_{1}=3$, $a_{n+1}=a_{n}^{3}$ for $n\geqslant 1$.
(Thus $a_{2}=3^{3}$, $a_{3}=(3^{3})^{3}$ and so on.)
\begin{questionparts}
\item What digit appears in the unit place of $a_{7}$?
\item Show that $a_{7}\geqslant 10^{100}$.
\item What is $\dfrac{a_{7}+1}{2a_{7}}$ correct to
two places of decimals? Justify your answer.
\end{questionparts}

Solution source

\begin{questionparts}
\item Notice that $a_n = 3^{3^{n-1}}$ in particular, $a_7 = 3^{3^6}$. Using Fermat's little theorem, we can see that $3^4 \equiv 1 \pmod{5}$ and so we need to figure out $3^6 \pmod{4}$, which is clearly $1$. Therefore $3^{3^6} \equiv 3^{4k+1} \equiv 3 \pmod{5}$. Therefore the units digit is $3$.

\item Notice that $3^5 > 100$ and $3^3 > 10$.
Therefore 
\begin{align*}
a_7 &= 3^{3^6} \\
&= (3^3)^{3^5} \\
&> 10^{3^5} \\
&> 10^{100}
\end{align*}
\item \begin{align*}
\frac{a_7+1}{2a_7} &= \frac12 + \frac1{2a_7} \\
&= 0.5 + 0.\underbrace{000\cdots}_{\text{at least }99\text{ zeros}} \\
&= 0.50
\end{align*}
Since $a_7 > 10^{100}, \, \frac{1}{2a_7} < 10^{-100}$
\end{questionparts}