A Checker Math Problem

[jaguar72]

Hello Gary Jenkins,

Congratulations!

You interpreted and answered the problem correctly.

Perhaps, I could have worded the latter half of the problem with greater clarity, by specifying that each of the squares within the checkerboard are congruent.

I might add that to my next revision.

The key to this problem was to recognize that the 4 corner squares are not to be counted twice.

For any given checkerboard, with y rows and y congruent squares in each row, then the number of these congruent squares which form the border or perimeter of the checkerboard can be solved by the the following:

4y - 4

For a standard 8 X 8 checkerboard, where y = 8, then the number of non-repeating border or perimeter squares is calculated as follows:

4 ( 8 ) - 4
32 - 4 = 28

ANSWER: CHOICE A (28)

None of the other choices work for the 10 X 10 or the 12 X 12 checkerboards.

Congratulations once again!

Best Wishes,

Dennis Cayton

Thanks, Mr. Cayton! My pleasure. The old K&E Log-Log Duplex Decitrig comes through again (well, actually, of course, I didn't have to use it for this problem but I did multiply the numbers on it just for the fun of it...).

Your posts/problems are quite excellent.

V/R,

Gary jenkins/jaguar72

[Danny_Alvarez]

Congrats to Gary for solving it correctly

I am not sure I would have interpreted correctly from the original wording Dennis.
I still enjoy these problems. Obviously i was sleeping on this one.... but as they say
"the early bird catches the worm"

cheers
Danny Alvarez