Find the greatest integer $k$

Relevant wiki: Binomial Theorem

$\begin{aligned} N = \displaystyle 1990^{1991^{1992}} + 1992^{1991^{1990}} &= (1991-1)^{1991^{1992}} + (1991+1)^{1991^{1990}} \\ &= \displaystyle \sum_{k=0}^{1991^{1992}} \dbinom{1991^{1992}}{k} (-1)^k 1991^{\left(1991^{1992}-k\right)} + \sum_{k=0}^{1991^{1990}} \dbinom{1991^{1990}}{k} 1991^{\left(1991^{1990}-k\right)} \\ & \small \color{#3D99F6} (\text{By binomial expansion}) \end{aligned}$

We notice that the powers of $1991$ in both the sum are decreasing, so to find the maximum value of $k$ for which $1991^k$ divides the above, we take the last two terms of each sum into consideration, let this be $S$

$\displaystyle \begin{aligned} S &= \left[ \dbinom{1991^{1992}}{1991^{1992} -1} 1991 - 1 \right] + \left[ \dbinom{1991^{1990}}{1991^{1990} -1} 1991 + 1 \right] \\ &= \displaystyle 1991 \left[ \dbinom{1991^{1992}}{1991^{1992} -1} + \dbinom{1991^{1990}}{1991^{1990} -1} \right] \\ &= \displaystyle 1991 \left( 1991^{1992} + 1991^{1990} \right) \\ &= \displaystyle 1991^{1993} + 1991^{1991} \end{aligned}$

Thus the smallest term in $N$ is $1991^{1991}$ which is divisible for $\boxed{k_{\text{ max}} = 1991}$ , i.e. by $1991^{1991}$ .

We have by Lift the exponent lemma:

$v_{1991}(1990^{1991^{1992}} + 1) = v_{1991}(1990+1)+ v_{1991}(1991^{1992}) = 1+ 1992 = 1993$

$v_{1991}(1992^{1991^{1990}} - 1) = v_{1991}(1992-1)+ v_{1991}(1991^{1990}) = 1 + 1990 = 1991$ .

Therefore $v_{1991}(1990^{1991^{1992}}+ 1992^{1991^{1990}}) = v_{1991}(1990^{1991^{1992}}+1 + 1992^{1991^{1990}} -1 ) = min(1991,1993) = \boxed{1991}$ .