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Here's a number sequence puzzle on INSC-2017:

132, 2646, 2128128, ?, 8482124821221022.5

I managed to get some progress but don't know how to complete it (progress deleted as comment request).

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  • $\begingroup$ I suggest not including your progress, whether it be in spoiler tags/quotes or not. Let us solve it as it is (and trust me, one of us will, but not me, probably)! :D $\endgroup$ – Mr Pie Sep 19 '18 at 10:37
  • $\begingroup$ @user477343 alright deleted now, good luck! $\endgroup$ – simonzack Sep 19 '18 at 10:39
  • $\begingroup$ Let the fun begin... $\endgroup$ – Mr Pie Sep 19 '18 at 10:42
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Although there is an aberration in the pattern, I believe the number is

42416241615.

The pattern is then

132, 2646, 2128128, 42416241615, 8482124821221022.5

Reasoning:

You have to double each number in the previous term and concatenate them. Then calculate the sum of all the digits in the current term and divide it by 2, and concatenate that number to the end.

Namely:

$2*2||1*2||2*2||8*2||1*2||2*2||8*2|| = 424162416$. Then $\frac{4+2+4+1+6+2+4+1+6}{2} = \frac{30}{2} = 15$, so the term is 42416241615. If we take this number, then $4*2||2*2||4*2||1*2||6*2||2*2||4*2||1*2||6*2||1*2||5*2|| = 84821248212210$. Then $\frac{8+4+8+2+1+2+4+8+2+1+2+2+1+0}{2} = \frac{45}{2} = 22.5$, so the term is 8482124821221022.5.

This works except for

the random 2 which starts 2128128, which should be a 4 (that term should instead be 4128129, then the next term ought to be 82416241818, and following that 16482124821621622.

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  • $\begingroup$ Can you explain how you arrive at the next number? I'm not quite sure thanks! $\endgroup$ – simonzack Sep 19 '18 at 12:48
  • $\begingroup$ @simonzack I’ve updated my answer. Funnily enough I think the question from the test isn’t as good as I think it could’ve been, but c’est la vie! $\endgroup$ – El-Guest Sep 19 '18 at 12:57
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    $\begingroup$ You're right, you figured out the last part, thanks! $\endgroup$ – simonzack Sep 19 '18 at 13:10

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