The last digit of #3^k# follows a cyclic pattern with a cycle length of #4#
#krarr"last digit of "3^k:color(white)("XX"){:
(0rarr1,1rarr3,2rarr9,3rarr7),
(4rarr1,5rarr3,6rarr9,7rarr7),
(8rarr1,9rarr3,10rarr9,11rarr7),
(...,...,...,...)
:}#
#7886 = 1971xx4+2#
So
#color(white)("XXX")3^7886 = (3^4)^1971xx3^2#
and
#color(white)("XXX")"the last digit of "3^7886 = 1xx"last digit of "3^2#
#color(white)("XXXXXXXXXXXXXXX")=9#
The digit in the units place of
#color(white)("XXX")2009 + 3^7886#
#color(white)("XXX")="last digit (units digit) of "(9+9)#
#color(white)("XXX")=8#
Perhaps the problem is in the meaning of the symbol on 2009.
viz. #|_underline(2009)#
I would normally understand this to be the "floor", the largest integer less than or equal to #2009#; but this would only normally be used for fractions, so perhaps it has some other meaning here.
