Without using a calculator, decide which of the numbers $33^{12}$, $63^{10}$, and $127^{8}$ is the smallest.
Let's observe that:

$$
\begin{aligned}
& 33^{12}>32^{12}=\left(2^{5}\right)^{12}=2^{60} \\
& 63^{10}<64^{10}=\left(2^{6}\right)^{10}=2^{60} \\
& 127^{8}<128^{8}=\left(2^{7}\right)^{8}=2^{56}
\end{aligned}
$$

Therefore, the largest of the numbers is $33^{12}$.

On the other hand, $\frac{127}{63}=2+\frac{1}{63}<2.1$. Therefore:

$$
\left(\frac{127}{63}\right)^{2}<2.1^{2}<7 \text { and }\left(\frac{127}{63}\right)^{4}<49<63 \Rightarrow 127^{4}<63^{5} \Rightarrow 127^{8}<63^{10}
$$

Therefore, the smallest of the three given numbers is $127^{8}$.