In the following \(6 \times 6\) array one can choose any \(k \times k\) subarray with \(1<k \leq 6 \) and add 1 to all its entries. Is it possible to perform the operation a finite number of times so that all the entries in the array are multiples of 3? \[ \begin{bmatrix} 2&0&1&0&2&0 \\ 0&2&0&1&2&0 \\ 1&0&2&0&2&0 \\ 0&1&0&2&2&0 \\ 1&1&1&1&2&0 \\ 0&0&0&0&0&0 \\ \end{bmatrix} \]

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TopNewestHint:Find an Invariant. What property must stay the same after any change?Log in to reply

Update: My original idea failed to work.

I would be interested in a solution.

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No, because following this Invariant principle we know that we will always have three different numbers x, x + 1 and x + 2. Since the differences between the numbers is 1 and 2, and not 3, this is not possible!

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How about considering that we just need to make the whole array's elements 0 (mod 3), so we wrote 2 as -1? Well that help here? Sorry I'm not much of a helper :(

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