Problem 775

［Solution］Obviously $n \geqslant 134$. On the other hand,
$$\begin{aligned} n^{5} & =133^{5}+110^{5}+84^{5}+27^{5} \\ & <133^{5}+110^{5}+(84+27)^{5} \\ & <3 \cdot 133^{5} \\ & <\frac{3125}{1024} \cdot 133^{5} \\ & =\left(\frac{5 \times 133}{4}\right)^{5} \\ & =\left(166-\frac{1}{4}\right)^{5}, \end{aligned}$$

Therefore, $n \leqslant 166$.
Since $a^{5}$ has the same last digit as $a$, the last digit of $n$ is the same as the last digit of $3+0+4+7$, which means the last digit of $n$ is $4$. Thus, the possible values of $n$ are $134, 144, 154$, or $164$.

Noting that 84 and 27 are both multiples of 3, and $133^{5}$ leaves a remainder of 1 when divided by 3, while $110^{5}$ leaves a remainder of 2 when divided by 3, it follows that $133^{5}+110^{5}+84^{5}+27^{5}$ is a multiple of 3, meaning $n$ is a multiple of 3.

Among $134, 144, 154, 164$, only 144 is a multiple of 3, so $n$ can only be 144.
Upon verification, $144^{5}=133^{5}+110^{5}+84^{5}+27^{5}$. Therefore, the value of $n$ is $144$.

[Solution] Clearly, $n \geqslant 134$. On the other hand,
$$\begin{array}{l} n^{5}= 133^{5}+110^{5}+84^{5}+27^{5} \\ <133^{5}+110^{5}+(84+27)^{5} \\ <3 \cdot 133^{5} \\ <\frac{3125}{1024} \cdot 133^{5} \\ =\left(\frac{5 \times 133}{4}\right)^{5} \\ =\left(166 \frac{1}{4}\right)^{5} \end{array}$$

Therefore, $n \leqslant 166$.
Since $a^{5}$ has the same last digit as $a$, the last digit of $n$ is the same as the last digit of $3+0+4+7$, which means the last digit of $n$ is $4$. Thus, the possible values of $n$ are $134, 144, 154$, or $164$.

Noting that 84 and 27 are both multiples of 3, and $133^{5}$ leaves a remainder of 1 when divided by 3, while $110^{5}$ leaves a remainder of 2 when divided by 3, it follows that $133^{5}+110^{5}+84^{5}+27^{5}$ is a multiple of 3, meaning $n$ is a multiple of 3.

Among $134, 144, 154, 164$, only 144 is a multiple of 3, so $n$ can only be 144.

Upon verification, $144^{5}=133^{5}+110^{5}+84^{5}+27^{5}$. Therefore, the value of $n$ is $144$.