Find the largest $k$ such that $1991^{k}$ divides $A=1990^{1991^{1992}}+1991^{1990^{1992}}+1992^{1991^{1990}}$.
We wish to use LTE, but the problem is that $1991=11 \cdot 181$. We will therefore have to evaluate the 11-adic and 181-adic valuations of $A$. Then, we imagine that the term $1991^{1990^{1992}}$ will not be limiting in this matter, so we will evaluate these valuations for $B=1990^{1991^{1992}}+1992^{1991^{1990}}=\left(1990^{1991^{2}}\right)^{1991^{1990}}+1992^{1991^{1990}}$. Thus, we need to verify that $1990^{1991^{2}}+1992$ is divisible by 1991, which is the case modulo 1991, so we can apply LTE to 11 and 181.

$$
v_{p}(B)=v_{p}\left(1990^{1991^{2}}+1992\right)+v_{p}\left(1991^{1990}\right)
$$

It remains to evaluate the valuation of $C=1990^{1991^{2}}+1992=1990^{1991^{2}}+1+1991$, so since $v_{p}\left(1990^{1991^{2}}+1\right)=v_{p}(1990+1)+v_{p}\left(1991^{2}\right)=3$, we deduce that $v_{p}(C)=1$, and $v_{p}(B)=1991$, hence for the two valuations $v_{11}(B)=v_{181}(B)=1991$, and indeed the term of $A$ in 1991 will not pose any problem, and we conclude that the maximum $k$ sought will be 1991.