Which is larger: $1985^{1986}$ or $1986^{1985}$?
Those who wrote approximate values for $\lg 1985$ and $\lg 1986$ instead of taking the logarithm after writing the inequality received 0 points. Since we estimate in inequalities, it does matter whether we use a lower or upper approximation for the given values.

Incorrect or unexplained solutions also received no points.

Since there were few correct solutions, we present a possible proof.

$$
\begin{aligned}
1986^{1985}= & (1985+1)^{1985}=1985^{1985}+\binom{1985}{1} \cdot 1985^{1984}+\ldots+ \\
& +\binom{1985}{1983} \cdot 1985^{2}+\binom{1985}{1984} \cdot 1985+1
\end{aligned}
$$

By the binomial theorem, we obtained the above sum with 1986 terms.

For $1 \leqq k \leqq 1983$, we estimate the $\binom{1985}{k} \cdot 1985^{1985-k}$ term from above.

$$
\begin{aligned}
& \binom{1985}{k} \cdot 1985^{1985-k}=\frac{1985 \cdot 1984 \cdot \ldots \cdot(1985-k+1)}{k!} \cdot 1985^{1985-k} \leqq \\
& \quad \leqq \underbrace{1985 \cdot \ldots \cdot(1985-k+1)}_{k \text { factors }} \cdot \underbrace{1985^{1985-k}}_{(1985-k) \text { factors }} \leqq 1985^{1985} .
\end{aligned}
$$

Finally:

$$
\binom{1985}{1984} \cdot 1985+1<1985^{2}+1985^{2}<1985^{1985}
$$

Thus:

$$
(1986)^{1985}<1985^{1985}+1983 \cdot 1985^{1985}+1985^{1985}=1985 \cdot 1985^{1985}=1985^{1986}
$$