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Puzzle: $$\begin{array}{}\boxed{3\;\,}&\boxed{150\;\;\;\,}&\boxed{12\;\;\;\,}\\\boxed{7\;\,}&\boxed{3164\;\,}&\boxed{56\;\;\;\,}\\\boxed{16}&\boxed{74112}&\boxed{272\;\;}\\\boxed{32}&\boxed{\text{?}\;\;\;\;\;\;\,}&\boxed{1024\,}\end{array}$$

We have to figure out the question mark.

My effort:

I have tried almost for half-an-hour, but can't seem to work out the rule being followed here. I tried
$$\begin{align} 7 - 3 &= 2^2\\ 16 - 7 &= 3^2\\ 32 - 16 &= 4^2 \end{align}$$

This was the only sequence I could figure out but it only works for the first column.

I need help figuring this out, along with a suitable explanation of the answer.

Note: This is not a homework question.

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  • $\begingroup$ From the looks of it, I would think that the first and last column are given and you derive the center column from the outside two. $\endgroup$ – Aggie Kidd Aug 14 '15 at 14:50
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    $\begingroup$ Interesting the left to the right appears to be in order 3*(3+1), 7*(7+1), 16*(16+1), then suddenly 32*32... $\endgroup$ – Going hamateur Aug 14 '15 at 14:51
  • $\begingroup$ @Goinghamateur Or if you look at it a different way, the ratios of 3rd/1st go 4, 8, 17, 32... $\endgroup$ – Patrick N Aug 14 '15 at 14:53
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My answer:

$\mathbf{1049088}$

The pattern I noticed was
$\text{middle column} = \text{right column}^2 + (\text{left column}\times 2^\text{row number})$

$2^1\times 3 + 12^2 = 6 + 144 = 150$
$2^2\times 7 + 56^2 = 28 + 3136 = 3164$
$2^3\times 16 + 272^2 = 128 + 73984 = 74112$
$2^4\times 32 + 1024^2 = 512 + 1048576 = 1049088$

My inspiration:

I divided 74112 by 272 and it came close to 272, so I though of squaring. Then I took the difference and said hey that looks like a multiple of 16, it was, I then applied the algorithm on all of them and the multiple trend of the difference from the square remained, and the 2, 4 ,8 pattern appeared. Leaving the assumption 16 would be the next difference multiplier.

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  • $\begingroup$ That was to fast for me :) Good work! $\endgroup$ – Wa Kai Aug 14 '15 at 14:56
  • $\begingroup$ +1 Ten seconds too fast for me :P I had gone through the squares and was about to run the last one, but you beat me to it! $\endgroup$ – Patrick N Aug 14 '15 at 14:57

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