Example 6 Determine $n^{5}=133^{5}+110^{5}+84^{5}+27^{5}$.
Solution: Since $n^{5} \equiv 3^{5}+0^{5}+4^{5}+7^{5} \equiv 3+4+7 \equiv$ $4(\bmod 10)$, the last digit of $n$ is 4. Clearly, the first digit of $n$ is 1. Further estimation: $n^{5}<2 \times 133^{5}+(84+$ $27)^{5}<3 \times 133^{5}<\left(\frac{5}{4}\right)^{5} \times 133^{5}$, so, $n<\frac{5}{4} \times$ $133<167$, hence $n$ can be $134,144,154,164$. Also, $n^{5}$ $\equiv 1^{5}+(-1)^{5} \equiv 3(\bmod 3)$,
thus $n=144$.