Yes
$1)21^{99}$
If a number has unit digit =1, this case applies
first, take unit digit and write in the unit place. __1
second, take the tens place of the number and multiplied to a unit digit of the power
in this example, $2*9=18$ take an only unit digit from this and write in the tens place.
we get last $2$ digits = 81
Similarly for $2)41^{9999}$
gives last $2$ digits $=61$
$3)31^{2019}$
gives last $2$ digits $=71$
$4)91^{2018!}$
$(91)^{@@@@...00}$ [10!=3,628,800,so 10! or greater number factorial gives atleast two zeros in the last $2$ digits]
first, take unit digit of the number and write in the unit place __ 1
second, take tense digit of the number and multiplied with a unit digit of the power.
$9*0=0$ write this in the tense place.
So,finally last $2$ digits $= 01$
I hope you understand!!