20030726, 16:06  #1 
Aug 2002
Ann Arbor, MI
433 Posts 
Checkerboard Problem

20030726, 18:45  #2 
Banned
"Luigi"
Aug 2002
Team Italia
2^{5}×149 Posts 
For the first puzzle, note that each domino has a black square and a white square, while the deleted corner squares are both white.
Each domino will cover exactly one black and one white square, leaving out (uncovered) a couple of black squares. Luigi 
20030726, 20:03  #3 
Jul 2003
5 Posts 
solution
ET nailed the first one.
The second problem is solved with very similar logic. Note that the 4 squares that are removed consist of 2 white and 2 black. This leaves 60 squares: 30 white, 30 black. Each T covers 4 squares. THus, 60/4 = 15 T's must be used. Since half a T cannot be used, we cannot use an equal number of each T. T1 covers 3W + 1B, T2 covers 1W + 3B. If we use A T1's and C T2's, then A + C = 15 => C +15  A. So, A(3W + B) + (15  A)(W + 3B) = 3AW + AB + 15W +45 B  AW  3B = (2A + 15)W + (A + 42)B = 60. (Note: W and B are not values. they are units.) (2A + 15)W = 30W because there are 30 white squares. thus, 2A = 15 => A=7.5 impossible because we have to use whole T's. Thus, the board cannot be covered. This proof could be a lot more compact, but I wanted to be fairly thorough. 
20030726, 21:39  #4 
Aug 2002
Ann Arbor, MI
433 Posts 
AllPrimesAreOdd You saw the solution in class, that's cheating. Let the other people have a go at it, first.

20030727, 01:11  #5 
Jul 2003
5 Posts 
I DID NOT
I did not just see the problem in class. I solved it.

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