There are two cases to consider.
A) The removed square is in an odd-numbered column (and row). In this case, the board is divided by that column and row into parts with an even number of columns, which can always be tiled by dominos, and the column the square is in, which has an even number of remaining squares that can also be tiled by dominos.
B) The removed square is in an even-numbered column (and row). In this case, the top row to the left of that column (including that column) can be tiled by dominos, as can the bottom row to the right of that column (including that column). The remaining untiled sections of the board have even numbers of rows, so can be tiled by dominos.
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Perhaps the shorter answer is that in an odd-sized board, the corner squares are the ones that there is one of in excess. Cutting out one that is of that color leaves an even number of squares, and equal numbers of each color. Such a board seems like it <em>ought</em> to be able to be tiled by dominos, but the above shows there is actually an algorithm for doing so.
The formula for arithmetic sequence is given by:
an=a+(n-1)d
where:
an=nth term
a=first term
n=number of terms
thus our first term will be:
a1=8
a2=8+(2-1)(-6)=2
a3=8+(3-1)(-6)=-4
a4=8+(4-1)(-6)=-10
a5=8+(5-1)(-6)=-16
a5=-16
thus the sequence will be:
8,2,-4,-10,-16...
Answer A seems to make the most sense to me<span />
recalling that d = rt, distance = rate * time.
we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?

so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.
then Wanda kicks in, rolling like a lightning at 16mph.
let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.
the distance both have travelled is the same "d" miles, reason why they "meet", same distance.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20Hector%26d%2612%26t%2B%5Cfrac%7B3%7D%7B2%7D%5C%5C%5B1em%5D%20Wanda%26d%2616%26t%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D%2812%29%5Cleft%28%20t%2B%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%5C%5C%5B1em%5D%20d%3D%2816%29%28t%29%20%5Cend%7Bcases%7D)

26 plus 27 is 53
hope I helped