Six circles, painted black or white, are in a row. At each step, a new line of six circles is drawn below and diagonally. The new circles of the new line are painted according to a simple rule that only depends on the previous line. See the figure below and discover what this rule is!

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-05.jpg?height=542&width=870&top_left_y=1104&top_left_x=670)

a) How will the circles of the seventh line be painted?

b) Will all the circles in a single line ever be painted black?

c) How will the circles of the line numbered 2014 be painted?

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Solution

a) Let's establish a simple rule to define the configuration of a row based on the configuration of the row immediately above it.

To determine the color of a ball, we must compare the colors of the two balls in the row above that are closest to it.
![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-23.jpg?height=338&width=844&top_left_y=997&top_left_x=486)

However, there is a problem! The last ball of a row has only one ball in the row above that is closest to it. "What should we do then to determine the color of the last ball of a given row?" To determine its color, we must compare the colors of the first and last ball of the row above. And apply the same rule described above.

Note that, with this rule, it is possible, starting from the coloring of the first row, to generate the coloring of the rows below one by one. The reader should verify that the rule described above really applies, starting with the first row colored as $P, B, P, B, B, P$ (here $P$ means "black" and $B$ means "white") and coloring the rows below until the sixth, repeating the drawing in the statement.

To generate the configuration of the seventh row, we use the fact that the configuration of the sixth row is given by $B, P, B, P, B, B$ and apply the rule above. The result is the coloring $P, P, P, P, B, B$ as we can see in the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-23.jpg?height=330&width=423&top_left_y=2066&top_left_x=698)
b) No! Suppose we had a row (say the $n$-th row) entirely colored black. For all the balls in this row to be colored black, it is necessary that the balls in the row above are colored alternately, that is, as $P, B, P, B, P, B$ or $B, P, B, P, B, P$.

Suppose the first case is true, that is, that the $(n-1)$-th row is colored as $P, B, P, B, P, B$. In this case, for the coloring of the first 5 balls of the $(n-1)$-th row to be as specified, it is necessary that the coloring of the $(n-2)$-th row be $B, P, P, B, B, P$ or $P, B, B, P, P, B$. But if the $(n-2)$-th row were colored in one of these ways, then the last ball of the $(n-1)$-th row would necessarily be colored black, which is not the case.

Now suppose the second case is true, that is, that the $(n-1)$-th row is colored as $B, P, B, P, B, P$. In this case, for the coloring of the first 5 balls of the $(n-1)$-th row to be as specified, it is necessary that the coloring of the $(n-2)$-th row be $B, B, P, P, B, B$ or $P, P, B, B, P, P$. But if the $(n-2)$-th row is colored in one of these ways, then the last ball of the $(n-1)$-th row would necessarily be colored white, which is not the case.
c) In item a), a rule was established to generate the coloring of the $n$-th row, provided the coloring of the $(n-1)$-th row is known. Through this rule, it was possible to generate the configuration of the seventh row from the configuration of the sixth row shown in the figure in the statement. The result obtained for the coloring of the seventh row is $P, P, P, P, B, B$. Applying the rule again, we obtain that the configuration of the eighth row will be given by $B, B, B, P, B, P$. Note that this is the same configuration as the third row. Thus, starting from the third row, the configurations will repeat in cycles of 6 rows. Therefore, the rows numbered $3, 9, 15, 21, 27, \ldots$, have the same configuration. Also, the rows numbered $3+k, 9+k, 15+k, 21+k, 27+k, \ldots$ have the same configuration as the row $k$ for each $k \in\{0, \ldots, 5\}$. Noting that $2014=3 \cdot 671+1$, we have that its coloring is the same as the row numbered $3+1$, that is, the row numbered 4. Therefore, the row numbered 2014 is colored as $B, P, B, B, B, P$.

Observation: The fact that the configurations repeat in cycles of 6 rows can also be used to solve item b). This fact allows us to know all the configurations that appear in this cycle. Listing all of them, we can observe that none of them is the one that contains all 6 black balls.