World Engineering Day Quiz!

1. for a 7 x 7 matrix 28 squares can be drawn that cover the center green cell.

2. The general formula for the number of squares that cover the center square is C=n(n+1)/2, where n is odd. So for n=99, C=4950.

1. For a 7x7 checkerboard, the number of squares is : Ns = 44
2. The general relation I establised is :

So, if n=99, Ns=83350.

Proof :

First question is easy... The second one will require more neurons...

Thanks rsprangh for your prompt reply. But I think you have been to fast.. There are in fact much more squares than 28 ! May be you want to retry?

Wow, big applause MikeLh , you found the solutions with a very clear and convincing development!

1) There are no squares covering the middle green square. There are eight squares bordering the green square that touch an edge of it with an additional 7x7-(8+1) or 40 other squares bordering it without touching it.

2) Assuming the same size squares, an extension to an NxN matrix (N odd) would still leave eight squares bordering the green square that touch its edge with an additional NxN-(8+1) other squares bordering it without touching it.

Screen shot to preserve maths:

Question 1:  How many squares, defined in that matrix, are covering the green cell?

This question is vague and nondescript and should be re-worded as:

Question 1:  How many tiles of MxM basic cells (M<=7) can be created that cover the green cell?

That said, the following is an update to my solution.

This quiz falls into the category of Checkerboard Tiling Problems where the checkerboard can be any NxN matrix of cells. N must be odd to create a center cell. I used a 9x9 table and visually created various sized tiles that covered the middle cell.  This led to the emergence of a sequence.

Tile Size             Number of Tiles

1x1                    1

2x2                    4

3x3                    9

4x4                    16

5x5                    25

6x6                    16

7x7                    9

8x8                    4

9x9                    1

Two expressions can be written and extended to any odd value of N to express the sum of this sequence.

I agree entirely with your comment on the posing of the question. And your answers are right according to my calculations (but two wrongs don't make a right). 

Thanks botha.
Our solutions only differ in the summation starting index. I chose 0 instead of 1 so at N=1 the range is 0=>0 rather than 1=>0. In either case, our results are correct and your effort in proving it by induction is appreciated.