11. (IMO-32 Shortlist) Find the largest positive integer $k$, such that
$$
1991^{k} \mid 1990^{19911992}+1992^{19911990} \text {. }
$$
11. First prove by mathematical induction:

Let odd number $a \geqslant 3$, for all positive integers $n$, we have
$$
(1+a)^{a^{n}}=1+S_{n} a^{n+1},
$$

where $S_{n}$ is an integer, and $a \times S_{n}$.
(1)
For $n=1$, we have
$$
(1+a)^{a}=1+C_{u}^{1} a+C_{u}^{2} a^{2}+\cdots+C_{u}^{a} a^{a}=1+a^{2}\left(1+C_{u}^{2}+C_{a}^{\beta} a+\cdots+a^{\alpha-2}\right) .
$$

Since $a$ is odd, $a \mid C_{a}^{2}$, thus
$a \mid C_{a}^{2}+C_{a}^{3} a+\cdots+a^{a-2}$, hence
$a \times S_{1}=1+C_{a}^{2}+\cdots+a^{a-2}$.
Therefore, for $n=1$, (1) holds.
Assume (1) holds for the natural number $n=k_{0}$, then
$$
\begin{aligned}
(1+a)^{a_{0}+1} & =\left[(1+a)^{k_{0}}\right]^{a} \\
& =\left(1+S_{k_{0}} a^{k_{0}+1}\right)^{a} \\
& =1+S_{k_{0}} a^{k_{0}+2}+C_{u}^{2} S_{k_{0}}^{2} a^{2 k_{0}+2}+\cdots+S_{k_{0}} a^{\left.a k_{0}+1\right)} \\
& =1+S_{k_{0}+1} a^{k_{0}+2},
\end{aligned}
$$

where $S_{k_{0}+1}=S_{k_{0}}+C_{a}^{2} S_{k_{0}}^{2} a_{0}^{k_{0}}+\cdots+S_{k_{0}}^{u} a^{a\left(k_{0}+1\right)-k_{0}-2}$.
By the induction hypothesis $a \times S_{k}$, thus $a \times S_{k_{0}+1}$.
Therefore, (1) holds for $n=k_{0}+1$.
That is, (1) holds for all natural numbers $n$.
Similarly, we can prove:
Let odd number $b \geqslant 3$, then for all positive integers $n$, we have
$$
(b-1)^{t^{n}}=-1+T_{n} b^{n+1} \text {, }
$$

where $T_{n}$ is an integer, and $b \times T_{n}$.
Using (1) and (2), we know that there exist integers $S$ and $T$, such that
$1991 \times S, 1991 \times T$, and
$$
1990^{1991^{1992}}+1992^{1991^{1990}}=T \cdot 1991^{1993}+S \cdot 1991^{1991}=1991^{1991}\left(T \cdot 1991^{2}+S\right) \text {. }
$$

Thus, the largest $k=1991$.