2006 Romanian NMO Problems/Grade 7/Problem 2: Difference between revisions
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A square of side <math>n</math> is formed from <math>n^2</math> unit squares, each colored in red, yellow or green. Find minimal <math>n</math>, such that for each coloring, there exists a line and a column with at least 3 unit squares of the same color (on the same line or column). | A square of side <math>n</math> is formed from <math>n^2</math> unit squares, each colored in red, yellow or green. Find minimal <math>n</math>, such that for each coloring, there exists a line and a column with at least 3 unit squares of the same color (on the same line or column). | ||
==Solution== | ==Solution== | ||
For <math> | For <math>n\leq6</math>, consider this coloring for a 6x6 board: | ||
<cmath>\begin{tabular}{|c|c|c|c|c|c|} | |||
\hline R&Y&G&R&Y&G \\ | |||
\hline G&R&Y&G&R&Y \\ | |||
\hline Y&G&R&Y&G&R \\ | |||
\hline R&Y&G&R&Y&G \\ | |||
\hline G&R&Y&G&R&Y \\ | |||
\hline Y&G&R&Y&G&R \\ | |||
\hline | |||
\end{tabular}</cmath> | |||
We can take the top <math>n</math>-by-<math>n</math> grid of this board as a coloring not satisfying the conditions. | We can take the top <math>n</math>-by-<math>n</math> grid of this board as a coloring not satisfying the conditions. | ||
Latest revision as of 01:48, 18 March 2009
Problem
A square of side 
is formed from 
unit squares, each colored in red, yellow or green. Find minimal , such that for each coloring, there exists a line and a column with at least 3 unit squares of the same color (on the same line or column).
Solution
For 
, consider this coloring for a 6x6 board:
![\[\begin{tabular}{|c|c|c|c|c|c|} \hline R&Y&G&R&Y&G \\ \hline G&R&Y&G&R&Y \\ \hline Y&G&R&Y&G&R \\ \hline R&Y&G&R&Y&G \\ \hline G&R&Y&G&R&Y \\ \hline Y&G&R&Y&G&R \\ \hline \end{tabular}\]](http://latex.artofproblemsolving.com/6/1/c/61cb61c72b0bdb96994aa6ed4cfd7746fd783793.png)
We can take the top -by- grid of this board as a coloring not satisfying the conditions.
For 
, we note that each row or column must have at least one color with 3 or more squares by the pigeonhole principle, so our answer is 7.
