10.80 Euler's conjecture was disproved by American mathematicians in 1960, who confirmed the existence of a positive integer $n$ such that $133^{5}+110^{5}+84^{5}+27^{5}=n^{5}$. Find the value of $n$.
(7th American Invitational Mathematics Examination, 1989)
［Solution］Obviously, from

we get
$$
\begin{array}{c}
n^{5}=133^{5}+110^{5}+84^{5}+27^{5}>133^{5}, \\
n \geqslant 134
\end{array}
$$

Next, we find the upper bound of $n$.
$$
\begin{aligned}
n^{5} & =133^{5}+110^{5}+84^{5}+27^{5} \\
& <133^{5}+110^{5}+(27+84)^{5} \\
& <3 \cdot 133^{5} \\
& <\frac{3125}{1024} \cdot 133^{5} \\
& =\left(\frac{5 \cdot 133}{4}\right)^{5} \\
& =\left(166 \frac{1}{4}\right)^{5}
\end{aligned}
$$

Therefore,
$$
n \leqslant 166 \text {. }
$$

Thus,
$$
134 \leqslant n \leqslant 166 \text {. }
$$

Since
$$
10 \mid a^{5}-a,
$$

the units digit of $a^{5}$ is the same as that of $a$.
Therefore, the units digit of $n$ is the same as that of $133+110+84+27$.
That is, the units digit of $n$ is 4.
We can search among $134,144,154,164$.
Since

we have
$$
\begin{array}{l} 
133 \equiv 1 \quad(\bmod 3), \\
110 \equiv 2 \quad(\bmod 3), \\
84 \equiv 0 \quad(\bmod 3), \\
27 \equiv 0 \quad(\bmod 3),
\end{array}
$$
$$
\begin{aligned}
n^{5} & =133^{5}+110^{5}+84^{5}+27^{5} \\
& \equiv 1^{5}+2^{5} \equiv 0 \quad(\bmod 3) .
\end{aligned}
$$

Among $134,144,154,164$, only
$$
144 \equiv 0 \quad(\bmod 3) .
$$

Therefore,
$$
n=144
$$