9
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The numbers below have been generated using a (quite) simple algorithm.

Can you spot the pattern and deduce the numbers in the next column and row?


$$ \begin{matrix} 1 & 2 & 2 & 4 & 2 & 4 & 4 & 8 & 2 & 4 & 4 & 8 & 4 & 8 & 8 & 16 & 2 & 4 & 4 & 8 & \cdots \\ 1 & 2 & 3 & 2 & 4 & 6 & 3 & 6 & 9 & 2 & 4 & 6 & 4 & 8 & 12 & 6 & 12 & 18 & 3 & 6 & \cdots \\ 1 & 2 & 3 & 4 & 2 & 4 & 6 & 8 & 3 & 6 & 9 & 12 & 4 & 8 & 12 & 16 & 2 & 4 & 6 & 8 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 2 & 4 & 6 & 8 & 10 & 3 & 6 & 9 & 12 & 15 & 4 & 8 & 12 & 16 & 20 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 2 & 4 & 6 & 8 & 10 & 12 & 3 & 6 & 9 & 12 & 15 & 18 & 4 & 8 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 3 & 6 & 9 & 12 & 15 & 18 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 3 & 6 & 9 & 12 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 3 & 6 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix} $$

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  • $\begingroup$ Quite easy to solve, but fun and a nice algorithm. (It must have been a challenge for you to work out how far to extend the table over to the right without either making the pattern obvious or making it appear far too simple for the lower rows!) $\endgroup$ – Rand al'Thor Sep 24 '16 at 16:53
8
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Start off with the following simple pattern:

$$ \begin{matrix} 1 & 2 & \cdots \\ 1 & 2 & 3 & \cdots \\ 1 & 2 & 3 & 4 & \cdots \\ 1 & 2 & 3 & 4 & 5 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix} $$

and multiply it by itself:

$$ \begin{matrix} 1 & 2 & 2 & 4 & \cdots \\ 1 & 2 & 3 & 2 & 4 & 6 & 3 & 6 & 9 & \cdots \\ 1 & 2 & 3 & 4 & 2 & 4 & 6 & 8 & 3 & 6 & 9 & 12 & 4 & 8 & 12 & 16 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 2 & 4 & 6 & 8 & 10 & 3 & 6 & 9 & 12 & 15 & 4 & 8 & 12 & 16 & 20 & \cdots \\ 1 & 2 & 3 & 4 & 5 & 6 & 2 & 4 & 6 & 8 & 10 & 12 & 3 & 6 & 9 & 12 & 15 & 18 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix} $$

Keep on applying the same process over and over again, multiplying the existing row by itself (the first element being 1, and 1 times itself being what's already there). In this way we can extend every row arbitrarily far to the right, and there are already infinitely many rows, so we get an infinite grid of numbers. Taking the first row as an example, it evolves as follows:

  • $1\:\:2$
  • $1\:\:2\:\:2\:\:4$
  • $1\:\:2\:\:2\:\:4\:\:2\:\:4\:\:4\:\:8\:\:2\:\:4\:\:4\:\:8\:\:4\:\:8\:\:8\:\:16$
  • etc.

So the elements of the next column in the grid given in the question are

$4,9,4,5,12,21,15,9,3,20,18$

and the elements of the next row are

$1,2,3,4,5,6,7,8,9,10,11,12,13,2,4,6,8,10,12,14,16$.

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  • 1
    $\begingroup$ Well, that's a very interesting way of looking at it! But my algorithm is actually even simpler :) $\endgroup$ – GOTO 0 Sep 24 '16 at 17:03
  • 1
    $\begingroup$ @GOTO0 My second matrix (the one just after the words "multiply it by itself")? I just double-checked, and all the numbers in that matrix match the ones in yours, as well as matching the algorithm as I've defined it. Admittedly my description could probably be made clearer, and I'm trying to think of a good way to do that. $\endgroup$ – Rand al'Thor Sep 24 '16 at 17:06
  • $\begingroup$ You're right @rand, seems like I was miscounting. Your description is clear enough, it's just that the method I'm using is different, although it yields the same results. $\endgroup$ – GOTO 0 Sep 24 '16 at 17:15
  • 3
    $\begingroup$ The algorithm of rand al'thor translates to: counting the rows from 2 and the columns from 0, the result is the product of all digits augmented by 1 of the column number in base row. So, write the column number in base row, add 1 to each digit and multiply them. $\endgroup$ – Florian F Sep 24 '16 at 18:34
  • $\begingroup$ @FlorianF That's in fact more on the line of what I had in mind. Good job anyway. $\endgroup$ – GOTO 0 Sep 24 '16 at 19:12

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