SMO 2013 Round 2 Q4

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? $\begin{bmatrix} 2&0&1&0&2&0 \\ 0&2&0&1&2&0 \\ 1&0&2&0&2&0 \\ 0&1&0&2&2&0 \\ 1&1&1&1&2&0 \\ 0&0&0&0&0&0 \\ \end{bmatrix}$

How to do this question? Please help. Thanks!

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Comments

Hint: Find an Invariant. What property must stay the same after any change?

Calvin Lin Staff - 5 years, 6 months ago

Update: My original idea failed to work.

I would be interested in a solution.

Calvin Lin Staff - 5 years, 6 months ago

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!

A Former Brilliant Member - 5 years, 6 months ago

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 :(

Timothy Wan - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$