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Consider the following problem:

There is an integer in each square of a $8 \times 8$ chessboard. In each move you can add $1$ to each of the integers in a smaller $4\times 4$ or $3 \times 3$ square. Can you always get a table with an entry divisible by $2$?

This problem can be solved by invariance. But the invariant quantity is quite difficult to invent. I tried; but I couldn't do it. The invariant quantity, it turned out was the fact that the sum of numbers in all the rows except the third and the sixth was invariant $mod\space 2$. After solving a couple of such problems (unsuccessfully), it started seeming to me that many such problems were hit-and-trial.

I don't think that this is what mathematics is about. There must be some underlying logic to the approach to the solution. In general, how can we approach such mathematical problems which appear to be pure hit-and-trial?

Gerard
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The invariant has to be something $\bmod 2$ since the question is asking whether the table has an entry divisible by $2$. It has to be something which is the same whether a $3\times 3$ or $4 \times 4$ square is chosen and whatever location it is in. Some sort of sum should work, but if all the squares are summed then $3\times 3$ and $4 \times 4$ squares add a different parity. Rows $3,6$ are a subset of the squares that have an odd number of squares intersecting $3\times 3$s and an even number intersecting $4\times 4$s.

If problems didn't involve any hit-and-trial and there was a method guaranteed to work every time they wouldn't be as much fun.

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    You are right. But some problems are just plain hit-and-trial. I don't think that's mathematics. Mathematics should involve creative solutions that arise from logical thought processes. A logical thought process does imply a repetitive one. I never said that there is a tried-and-tested method that will guarantee that a problem is solved. – Gerard Jul 27 '13 at 07:57