20) In each of the following chains of inequalities, the same five numbers appear; which of them is completely true?
(A) $2^{3^{2}}<2^{3^{3}}<3^{2^{2}}<3^{2^{3}}<3^{3^{2}}$
(B) $3^{2^{2}}<3^{2^{3}}<2^{3^{2}}<2^{3^{3}}<3^{3^{2}}$,
(C) $3^{2^{2}}<3^{2^{3}}<2^{3^{2}}<3^{3^{2}}<2^{3^{3}}$,
(E) $3^{2^{2}}<2^{3^{2}}<3^{2^{3}}<3^{3^{2}}<2^{3^{3}}$.
(D) $3^{2^{2}}<2^{3^{2}}<3^{2^{3}}<2^{3^{3}}<3^{3^{2}}$,
20. The answer is $(\mathbf{E})$. First solution. $\mathbf{2}^{\mathbf{3}^{\mathbf{3}}}=2^{27}>2^{18}=4^{9}>3^{9}=\mathbf{3}^{\mathbf{3}^{\mathbf{2}}}>3^{8}=\mathbf{3}^{\mathbf{2}^{\mathbf{3}}}=\left(3^{2}\right)^{4}=$ $9^{4}>8^{4}=\left(2^{3}\right)^{4}=2^{12}>2^{9}=\mathbf{2}^{\mathbf{3}^{\mathbf{2}}}=4^{4}>3^{4}=\mathbf{3}^{\mathbf{2}^{\mathbf{2}}}$.

Second solution. The numbers in the various chains of inequalities are (we only report the results of calculations that can be done without a calculator in a reasonable time):

$$
\begin{aligned}
& 3^{2^{2}}=3^{4}=81 \\
& 2^{3^{2}}=2^{9}=512 \\
& 3^{2^{3}}=3^{8}=(81)^{2}=7371 \\
& 3^{3^{2}}=3^{9}=3^{8} \cdot 3=7371 \cdot 3=22113 \\
& 2^{3^{3}}=2^{27}=2^{10} \cdot 2^{10} \cdot 2^{7}=(1024)^{2} \cdot 2^{7}>\left(10^{3}\right)^{2}=10^{6}>22113 .
\end{aligned}
$$

Therefore, the correct chain of inequalities is (E): $3^{2^{2}}<2^{3^{2}}<3^{2^{3}}<3^{3^{2}}<2^{3^{3}}$.

Finally, observe that (A), (B), and (D) are incorrect because the number $2^{3^{3}}$ does not appear as the last term on the right. (C) is incorrect because it contains the inequality $3^{2^{3}}=7371<2^{3^{2}}=512$, which is false.