Value of k

Well, I can prove that the expn is not divisible by 1991;
This can be done like this

1992-(1) is div by 1991
therefore 1992≡1(mod1991)
=> 1992¹⁹⁹¹ ≡1(mod1991)
=>1992^19911990≡1(mod1991)...1

similarly 1990-(-1) is div by 1991
proceeding in the same way , we get
1990^19911992≡1(mod 1991)....2
Adding 1 and 2
1990^{1991 1992} +1992^19911990≡2(mod1991)
If this method is correct, how can we conclude that 0 is the highest possible value?

@Ranadeep Roy:

1991¹⁹⁹² is odd.Therefore, 1990^19911992≡(-1)^{odd number}≡-1(mod 1991)

Hence 1990^{1991 1992} +1992^19911990≡1-1=0(mod1991)

So k has to be at least 1.

Expand it using binomial theorem.
You'll find that all the terms are divisible by 1991¹⁹⁹¹.

Hence k = 1991¹⁹⁹¹