By the binomial theorem is easy to prove that if $ x \equiv 1 (mod m^k)$ then $ x^{m^n} \equiv 1 (mod m^{k+n})$ for every integers positive $ x, m$, and if $ m$ is odd and $ x$ satisfying $ x \equiv -1 (mod m^k)$ then $ x^{m^n} \equiv -1 (mod m^{k+n})$. Thus $ 1992^{1991^{1990}} \equiv 1 (mod 1991^{1991})$ and $ 1990^{1991^{1992}} \equiv -1 (mod 1991^{1993})$. Hence, the highest degree is $ 1991$.

unless I am mistaken, this is a simple application of lifting the exponent. $ 1991 = 11 \times 181$ let $ S = 1990^{1991^{1992}} + 1992^{1991^{1990}} = (1990^{1991^{2}})^{1991^{1990}} - ( - 1992)^{1991^{1990}}.$ then the highest degree of $ 11$ dividing $ S$ equals $ 1990 + 1 = 1991$ since $ 11\|(1990^{1991^{2}}) - ( - 1992) = 1991 [(\sum_{i = 0}^{1991^{2} - 1}( - 1990)^{i}) + 1].$ the same thing goes with $ 181$ and we are done

More simple! We let $1991=a$ then $a$ is a prime. By Lifting The Exponent lemma \[\begin{aligned} v_a \left( (a-1)^{a^{a+1}}+ (a+1)^{a^{a-1}} \right) & = v_a \left( (a-1)^{a^2})^{a^{a-1}}+ (a+1)^{a^{a-1}} \right) \\ & = v_a \left( (a-1)^{a^2}+a+1 \right)+ v_a(a^{a-1}) \\ & = a-1+v_a \left( (a-1)^{a^2}+a+1 \right) \end{aligned} \] We have $v_a \left( (a-1)^{a^2}+1 \right)= v_a(a)+v_a(a^2)=3$ so $v_a \left( (a-1)^{a^2}+a+1 \right) =1$. Thus, $v_a \left( (a-1)^{a^{a+1}}+ (a+1)^{a^{a-1}} \right) = a$. We obtain $\max k =a= \boxed{1991}$.

orl wrote: Find the highest degree $ k$ of $ 1991$ for which $ 1991^k$ divides the number \[ 1990^{1991^{1992}} + 1992^{1991^{1990}}.\] Note that $1990^{1991^{1992}}=(1990^{1991^2})^{1991^{1990}}$. Hence by LTE, \begin{align*}v_{1991}(1990^{1991^{1992}}+1992^{1991^{1990}}) & =v_{1991}((1990^{1991^2})^{1991^{1990}}+1992^{1991^{1990}})\\ &=v_{1991}(1990^{1991^2}+1992)+v_{1991}(1991^{1990)})\\ &=v_{1991}(1990^{1991^2}+1992)+1990.\end{align*} It remains to find $v_{1991}(1990^{1991^2}+1992)$. We re-write this as $v_{1991}(1990^{1991^2}+1+1992-1)=\min\{v_{1991}(1990^{1991^2}),v_{1991}(1992-1)\}$. The second one is $1$, so we check $v_{1991}(1990^{1991^2}+1)=v_{1991}(1991)+v_{1991}(1991^2)=3>1$. It follows that $v_{1991}(1990^{1991^2}+1992)=1$. Thus, \begin{align*}v_{1991}(1990^{1991^{1992}}+1992^{1991^{1990}}) & =v_{1991}(1990^{1991^2}+1992)+1990\\ &=1+1990\\ &=\boxed{1991}.\end{align*}

shinichiman wrote: More simple! We let $1991=a$ then $a$ is a prime. Umm nope since $1991=11 \times 181$.

I did this by letting $1990 = 1991 -1 $ and $1992 = 1991 +1 $ and then expanding using the binomial theorem. The "ones" cancel out and we are left with giant powers of $1991$. Observing the second to last term tells us that the minimal degree of a $1991$ is indeed $1991$.

shinichiman wrote: More simple! We let $1991=a$ then $a$ is a prime. By Lifting The Exponent lemma \[\begin{aligned} v_a \left( (a-1)^{a^{a+1}}+ (a+1)^{a^{a-1}} \right) & = v_a \left( (a-1)^{a^2})^{a^{a-1}}+ (a+1)^{a^{a-1}} \right) \\ & = v_a \left( (a-1)^{a^2}+a+1 \right)+ v_a(a^{a-1}) \\ & = a-1+v_a \left( (a-1)^{a^2}+a+1 \right) \end{aligned} \]We have $v_a \left( (a-1)^{a^2}+1 \right)= v_a(a)+v_a(a^2)=3$ so $v_a \left( (a-1)^{a^2}+a+1 \right) =1$. Thus, $v_a \left( (a-1)^{a^{a+1}}+ (a+1)^{a^{a-1}} \right) = a$. We obtain $\max k =a= \boxed{1991}$. 1991 isn't a prime.

o0f bash Solution Note that $$1990^{1991^{1992}}+1992^{1991^{1990}}=(1990^{1991^{2}})^{1991^{1990}}+1992^{1991^{1990}}.$$Since $1991$ may be prime factorized as $11\cdot 181$, $$v_{1991}((1990^{1991^{2}})^{1991^{1990}}+1992^{1991^{1990}})=\min(v_{11}((1990^{1991^{2}})^{1991^{1990}}+1992^{1991^{1990}}),v_{181}((1990^{1991^{2}})^{1991^{1990}}+1992^{1991^{1990}})).$$Since $$1990^{1991^{2}}+1992\equiv (-1)^{1991^{2}}+1\equiv 0 \pmod{11}$$and $1991^{1990}\equiv 1\pmod{2}$, the Lifting the Exponent Lemma yields $$v_{11}((1990^{1991^{2}})^{1991^{1990}}+1992^{1991^{1990}})=v_{11}(1990^{1991^{2}}+1992)+v_{11}(1991^{1990}).$$Note that $$v_{11}(1990^{1991^{2}}+1992)=v_{11}((181\cdot 11-1)^{1991^{2}}+181\cdot 11 + 1)=$$$$v_{11}((-1+\tbinom{1991^2}{1}\cdot 181 \cdot 11 - \tbinom{1991^2}{2}\cdot 181^2 \cdot 11^2 +...)+181\cdot 11 +1)=v_{11}((1991^2+1)\cdot 181\cdot 11) = 1.$$Note that $v_{11}(1991^{1990})=1990$, thus $v_{11}((1990^{1991^{2}})^{1991^{1990}}+1992^{1991^{1990}})=1991.$ $\square$ Since $$1990^{1991^{2}}+1992\equiv (-1)^{1991^{2}}+1\equiv 0 \pmod{181}$$and $1991^{1990}\equiv 1\pmod{2}$, the Lifting the Exponent Lemma yields $$v_{181}((1990^{1991^{2}})^{1991^{1990}}+1992^{1991^{1990}})=v_{181}(1990^{1991^{2}}+1992)+v_{181}(1991^{1990}).$$Note that $$v_{181}(1990^{1991^{2}}+1992)=v_{181}((181\cdot 11-1)^{1991^{2}}+181\cdot 11 + 1)=$$$$v_{181}((-1+\tbinom{1991^2}{1}\cdot 181 \cdot 11 - \tbinom{1991^2}{2}\cdot 181^2 \cdot 11^2 +...)+181\cdot 11 +1)=v_{181}((1991^2+1)\cdot 181\cdot 11) = 1.$$Note that $v_{181}(1991^{1990})=1990$, thus $v_{181}((1990^{1991^{2}})^{1991^{1990}}+1992^{1991^{1990}})=1991.$ $\square$ Thus $k=v_{1991}((1990^{1991^{2}})^{1991^{1990}}+1992^{1991^{1990}})=\min(1991,1991)=1991$, as desired. $\blacksquare$

LEMMA:- For every odd number $a\ge 3$ and integer $n\ge 0,$ the following holds. $$a^{n+1}||(a+1)^{a^n} - 1, a^{n+1}|| (a-1)^{a^n} + 1$$. These lemma can be proved by induction on $n$. The problem is trivial from here.

The answer is $\boxed{1991}$. We can rewrite the expression as $$\nu_{1991}(1990^{1991^{1992}} + 1992^{1991^{1990}}) = \nu_{1991}([1990^{1991^2}]^{1991^{1990}} + 1992^{1991^{1990}}$$$$= \nu_{1991}([1990^{1991^2}]^{1991^{1990}} - (-1992^{1991^{1990}})).$$By the LTE lemma, this is equal to $$1990 + \nu_{1991}(1990^{1991^2} + 1992).$$The finishing step is to realize that $$\nu_{1991}(1990^{1991^2} + 1) = 1 + 2 = 3,$$which implies that the second $\nu_{1991}$ must be equal to exactly 1, hence the result. Remark. Note that 1991 is not prime, so some of our early manipulations might be handwavy directly. However, we can simply replace 1991 by the product of its two distinct prime factors then run the algorithm similarly.

The answer is $\boxed{1991}$. Notice that $1991 \mid 1990^{1991^{1992}} + 1992^{1991^{1990}}$ and $1981=11*181$ So,$11,181 \mid 1990^{1991^{1992}} + 1992^{1991^{1990}}$ and moreover $11,181 \mid 1990+1992$ and thus we are ready to use LTE Applying $\nu_{11}$ both side,we get $$k \leq \nu_{11}(1990^{1991^{1992}} + 1992^{1991^{1990}}=(1990^{1991^{2}})^{1991^{1990}} +1992^{1991^{1990}})$$$$=\nu_{11}(1990^{1991^{2}}+1992)+\nu_{11}(1991^{1990})=\nu_{11}(1990^{1991^{2}}+1992)+1990=1991$$Similarly, $$\nu_{181}(1990^{1991^{2}}+1992)+\nu_{181}(1991^{1990})=\nu_{181}(1990^{1991^{2}}+1992)+1990=1991$$which implies $\boxed{k=1991}$

If Alexander Remorov is right,this is N5.Also I don't think that the above solution is correct. ISL 1991 N5 wrote: Find the highest degree $ k$ of $ 1991$ for which $ 1991^k$ divides the number \[ 1990^{1991^{1992}} + 1992^{1991^{1990}}.\] First we find $v_{11}(1990^{{1991}^2}+1992)$.Alternately we will prove that $v_{121}(1990^{{1991}^2}+1992)=0$.We will be using Fermats little theorem and Eulers theorem repeatedly,so we will not state the time of their usages. \begin{align*}&(1990^{1991})^{1991}+1992 \equiv (54^{1991})^{1991}+1992 \pmod{121} \\ &\implies (54^{11})^{1991}+1992 \equiv (54^{1991})^{11}+1992 \pmod{121} \\ &\implies 54^{{11}^{11}}+1992 \equiv 54^{11}+1992 \pmod{121} \\ &\implies (54^2)^5.54+1992 \equiv 12^5.54+1992 \pmod{121} \\ &\implies 1728.144.54+56 \equiv 34.23.54+56 \pmod{121} \\ &\implies 782.54+56 \equiv 56.54+56 \equiv 56.55 \ne 0 \pmod{121} \end{align*}Again by standard modulo work we get $11$ divides the expression $1990^{{1991}^2}+1992$ and thus $v_{11}$ of the expression is $1$. Now we will use L.T.E $$v_{11}(1990^{{1991}^{1992}}+1992^{{1991}^{1990}})=v_{11}((1990^{{1991}^2})^{{1991}^{1990}}+1992^{{1991}^{1990}})=v_{11}(1990^{{1991}^2}+1992)+1990=1991$$Similarly we get that $$v_{181}(1990^{{1991}^{1992}}+1992^{{1991}^{1990}}) >1990$$Since $1991=11.181$ we can simply ignore the $v_{181}$ condition and write $k_{\text{max}}=1991$ $\blacksquare$

We claim the answer is $1991.$ Notice $1991=11\cdot 181$ and $k\ge 1.$ Hence, by LTE, $$\nu_{11}\left(\left(1990^{1991^2}\right)^{1991^{1990}} + 1992^{1991^{1990}}\right)=\nu_{11}(1990^{1991^2}+1992)+\nu_{11}(1991^{1990}).$$Notice $$(1991-1)^{1991^2}+1992\equiv -1+1992\equiv 1991\pmod{1991^2}$$so $\nu_{11}(1990^{1991^2}+1992)=1$ and $k\le 1991.$ Similarly for $181,$ we can conclude that $181^{1990+1}$ divides our number, so $k=1991$ is achievable. $\square$

bruh Let's first do a bit of manipulation so that exponents match: $$1990^{1991^{1992}}=1990^{1991^2\cdot 1991^{1990}}=(1990^{1991^2})^{1991^{1990}}.$$Note that $1992+1990^{1991^2}$ is a multiple of 11 since the first term is 1 mod 11 and the second is -1 mod 11, so by LTE we have $$v_{11}(1992^{1991^{1990}}-(-1990^{1991^2})^{1991^{1990}})$$$$=v_{11}(1992+1990^{1991^2})+v_{11}(1991^{1990})=v_{11}(1992+1990^{1991^2})+1990.$$We can also compute that $$1992+1990^{1991^2}\equiv 56+54^{1991^2}\equiv 56+54^{11^2}\equiv 56+54^{11}\equiv 55\pmod {121},$$so $v_{11}(1992+1990^{1991^2})=1$ and the $v_{11}$ of the entire thing is $1991$. We proceed similarly with $$v_{181}(1992^{1991^{1990}}-(-1990^{1991^2})^{1991^{1990}})$$$$=v_{181}(1992+1990^{1991^2})+v_{181}(1991^{1990})=v_{181}(1992+1990^{1991^2})+1990.$$We do some more tedious calculation to see that $v_{181}(1992+1990^{1991^2})=1$, so this is also 1991. Therefore, the answer is $\min(1991,1991)=1991.$

Let $x=1990^{1991^2}, y=1992$. Our expression can be rewritten as $$A=x^{1991^{1990}}-(-y)^{1991^{1990}}.$$Note that (using $p=11, 181$) all the conditions for LTE are met. Then we have $$\nu_{11}(A)=1+\nu_{11}(1991^{1990})=1991,$$and similarly $\nu_{181}(A)=1991$, so the largest such $k$ is $1991$.

Cool, but I just wanted to mention out that after getting $\nu_{11}(A)=1991$, we don't actually need to check it for $\nu_{181}(A)$ since it will also be $\ge 1991$ as $181 \mid 1990^{1991^2}+1992$. So we automatically get $k=1991$. Many of the above posts actually calculated it.

Two main realizations here: - We should make the power the same to use LTE - LTE can be used even in addition by negating the second term With this in mind, $\nu_{11}(A^{1991^{1990}} + B^{1991^{1990}}) = \nu_{11}(A + B) + \nu_{11}(1991^{1990}) = 1 + 1990 = 1991$ ($A = 1990^{1991^2}, B=1992$), $\nu_{11}(A + B) = 1$ by binomial expansion. Now, we don't even have to compute $\nu_{181}(A^{1991^{1990}} + B^{1991^{1990}})$, as we know it will be $\ge 1991$. Regardless, our final answer is $\boxed{1991}$.

We can rewrite the given as $1990^{1991^{1992}} + -1^{1991^{1992}} + 1992^{1991^{1990}} + 1^{1991^{1990}}$. We can use LOE to see that $v_p(1990^{1991^{1992}} + -1^{1991^{1992}}) = 1 + 1992$ and $v_p(1992^{1991^{1990}} + 1^{1991^{1990}}) = 1 + 1990$. For $p = 11, 181$. Since $v_p(x+y) = \min(x, y)$ if $x \neq y$ and $1991 = 11 \cdot181$ and $11$ and $181$ are prime, we can see our answer of $\boxed{1991}$. $\blacksquare$

Just write $1991=11 \times 181$ and use LTE on both of those numbers.

S.Das93 wrote: Just write $1991=11 \times 181$ and use LTE on both of those numbers. yea i know I just needed to go right then

Math4Life7 wrote: We can rewrite the given as $1990^{1991^{1992}} + -1^{1991^{1992}} + 1992^{1991^{1990}} + 1^{1991^{1990}}$. We can use LTE to see that $v_p(1990^{1991^{1992}} + -1^{1991^{1992}}) = 1 + 1992$ and $v_p(1992^{1991^{1990}} + 1^{1991^{1990}}) = 1 + 1990$. For $p = 11, 181$. Since $v_p(x+y) = \min(x, y)$ if $x \neq y$ and $1991 = 11 \cdot181$ and $11$ and $181$ are prime, we can see our answer of $\boxed{1991}$. $\blacksquare$

The answer is $1991$. We implement LTE. First express $$1990^{1991^{1992}} + 1992^{1991^{1990}} = (1990^{1991^2})^{1991^{1990}} + (1992)^{1991^{1990}}$$ Note that $1991^{1990}$ is odd and $11 \mid 1990^{1991^2} + 1992$ but $11 \nmid 1990^{1991^2}$ and $11 \nmid 1992$ so we can apply LTE. It follows that $$v_{11}((1990^{1991^2})^{1991^{1990}} + (1992)^{1991^{1990}}) = v_{11}(1990^{1991^2}+1992) + v_{11}(1991^{1990}) = v_{11}(1990^{1991^2}+1992) + 1990$$ With some mod bashing we see that $v_{11}(1990^{1991^2}+1992) = 1$ so $v_{11}((1990^{1991^2})^{1991^{1990}} + (1992)^{1991^{1990}}) = 1991$. Note we can follow a similar process to get that $v_{181}((1990^{1991^2})^{1991^{1990}} + (1992)^{1991^{1990}}) \geq 1991$ so the answer is $1991$.

We claim the answer is $1991=11\cdot 181$ which clearly works. We now prove this is the maximum. Let $x=1990^{1991^2}$ and $y=1992$. Looking at $x+y$, we can see that it is: \[1991^{1991^2}-\dots+\binom{1991^2}{1}\cdot 1991-1+1991+1\equiv 1991\cdot 2\pmod{1991^2}.\] Note that $x+y\equiv 0\pmod{11}$. By LTE: \begin{align*} v_{11}\left(1990^{1991^{1992}} + 1992^{1991^{1990}}\right)&=v_{11}(x+y)+1990\\ &=1991.\\ \end{align*}Similarly, $v_{181}(x+y)=1991$, so we have our desired answer.

Let $A= {1990^{1991}}^2$ and $B=1992$. We want to find \[\nu_{1991}(A^{1991^{1990}}+B^{1991^{1990}})\] Notice for $p \in \{11,181\}$, we have \[\nu_p(A^{1991^{1990}}+B^{1991^{1990}})=\nu_p(A+B)+\nu_p(1991^{1990}) = 1991\] so we conclude that the answer is $\boxed{1991}$.

First, take $v_{11}$ of the expression, and note that by LTE, \[ v_{11}(1990^{1991^{1992}} + 1992^{1991^{1990}} = v_{11}((1990^{1991^2})^{1991^{1990}} + 1992^{1991^{1990}}) = v_{11}(1990^{1991^2} + 1992) + 1990 \]Note that $v_{11}(1990^{1991^2} + 1992^{1991^2}) = v_{11}(1991*2) + v_{11}(1991^2) = 3$. However, we also have that \[ v_{11}(1992^{1991^2} - 1992) = v_{11}(1992^{1990 \cdot 1992} - 1) = v_{11}(1991) + v_{11}(1990 \cdot 1992) = 1 \]Thus, $v_{11}(1990^{1991^2} + 1992) = 1$, and we find that $v_{11}$ of the expression is $1991$. A similar argument with $v_{181}$ also holds, so the answer is $\boxed{1991}$. $\blacksquare$

Note that $1991 = 181 \cdot 11$. Suppose $p$ is one of these two primes. Using LTE, we get \[v_p \left((1990^{1991^2})^{1991^{1990}} + (1992)^{1991^{1990}}\right) = v_p(1990^{1991^2}+1992) + 1990.\] From Binomial Theorem, we know \begin{align*} 1990^{1991^2}+1992 &= (1991-1)^{1991^2}+1992 \\ &\equiv 1991 \cdot \binom{1991^2}{1}-1+1992 \\ &\equiv 1991 \pmod{1991^2} \end{align*} Hence $v_p(1990^{1991^2}+1992)=1$, so our ansswer is $\boxed{1991}$.

Rewrite $1990^{1991^{1992}} + 1992^{1991^{1990}} = \left(1990^{1991^{1992}} + 1\right) + \left(1992^{1991^{1990}}-1\right)$. Now observe that $1991 = 11*181$. From LTE, $\nu_{11}\left(1990^{1991^{1992}} + 1\right) = \nu_{11}\left(1991\right) + \nu_{11}\left(1991^{1992}\right) = 1993$, $\nu_{11}\left(1992^{1991^{1990}} - 1\right) = \nu_{11}\left(1991\right) + \nu_{11}\left(1991^{1990}\right) = 1991$. Therefore, $\nu_{11}\left(\left(1990^{1991^{1992}} + 1\right) + \left(1992^{1991^{1990}}-1\right)\right) = 1991$. From similar calculations, $\nu_{181}\left(\left(1990^{1991^{1992}} + 1\right) + \left(1992^{1991^{1990}}-1\right)\right) = 1991$. Therefore, the answer is $\boxed{k = 1991}$. $\square$

$v_{11}(1990^{1991^{1992}} + 1992^{1991^{1990}})=v_{11}((1990^{1991^2})^{1991^{1990}} + 1992^{1991^{1990}})=v_{11}(1990^{1991^2}+1992)+v_{11}(1991^{1990})=1+1990=1991$ Use the same logic for $181$ to get that it equals $v_{11}(1990^{1991^2}+1992)+1990 > 1+1990 = 1991$, since $1990^{1991^2}+1992$ divides $181$, and it might divide $181^2$, but that does not matter, since it does not divide $121^2$ So the answer is $\boxed{1991}$

Max D.R. wrote: By the binomial theorem is easy to prove that if $ x \equiv 1 (mod m^k)$ then $ x^{m^n} \equiv 1 (mod m^{k+n})$ for every integers positive $ x, m$, and if $ m$ is odd and $ x$ satisfying $ x \equiv -1 (mod m^k)$ then $ x^{m^n} \equiv -1 (mod m^{k+n})$. Thus $ 1992^{1991^{1990}} \equiv 1 (mod 1991^{1991})$ and $ 1990^{1991^{1992}} \equiv -1 (mod 1991^{1993})$. Hence, the highest degree is $ 1991$. Thanks for eliminating some of the LTE corruption from this page. Initially I thought you got this motto from "nothing", but maybe it will make sense after some progress on the problem.

We have that 1991 = 11.181. Let p be 11 or 181. We can now change the statement to $1990^{1991^{1992}} + 1^{1991^{1992}} + 1992^{1991^{1990}} - 1^{1991^{1990}}$. By LTE we get that $\nu_p(1992^{1991^{1990}} - 1^{1991^{1990}}) = \nu_p(1991) + \nu_p(1991^{1990}) = 1991$ and $\nu_p(1990^{1991^{1992}} + 1^{1991^{1992}}) = \nu_p(1991) + \nu_p(1991^{1992}) = 1993$ $\Rightarrow$ $\nu_p(1990^{1991^{1992}} + 1992^{1991^{1990}}) = \min\left\{\nu_{p}(\nu_p(1992^{1991^{1990}} - 1^{1991^{1990}})),\nu_{p}(\nu_p(1990^{1991^{1992}} + 1^{1991^{1992}}))\right\} = 1991$ $\Rightarrow$ the number we searched is k = 1991.

We split $1991$ into $11, 181$. Then note that $1990^{1991^{1992}}+1992^{1991^{1990}}=(1990^{1991^2})^{1991^{1990}}+1992^{1991^{1990}}$. Then by LTE $\nu_{11}((1990^{1991^2})^{1991^{1990}}+1992^{1991^{1990}})=\nu_{11}(1990^{1991^2}+1992)+\nu_{11}(1991^{1990})=\nu_{11}(1990^{1991^2}+1992)+1990$. We can check easily that $11 \mid 1990^{1991^2}+1992$. Now, we calculate $1990^{1991^2}+1992 \mod{11^2}$. Since $1991^2 \equiv 11^2 \equiv 11 \mod{\varphi(11^2)}$, we have $54^{11}+56 \equiv (55-1)^{11}+56 \equiv 11(55)-1+56 \equiv 55 \mod{11^2}$. Thus, $\nu_{11}=1991$. We can easily check that $\nu_{181}$ is at least $1991$ similarly, so $k=1991$.

Let $N$ denote the integer in question; we have \[N = \left(1990^{1991^2}\right)^{1991^{1990}} + 1992^{1991^{1990}}.\]Noting that $1991 = 11 \cdot 181$, by LTE we obtain \[\nu_{11}(N) = \nu_{11}\left(1990^{1991^2} + 1992\right) + \nu_{11}(1991^{1990}) = 1 + 1990 = 1991,\]since \[\nu_{11}\left(1990^{1991^2} + 1^{1991^2}\right) = \nu_{11}(1991) + \nu_{11}(1991^2) = 3\]and \[\nu_{11}\left(1990^{1991^2} + 1992\right) = \nu_{11}\left(\left(1990^{1991^2} + 1\right) + 1991\right) = \min(3, 1) = 1.\]Similarly, $\nu_{181}(N) = 1991$ for the exact same reasons. Thus the largest such integer $k$ is $1991$. $\square$