3968, 63, 8, 3.

I'm lost here, thought it had something to do with square roots but my guess was 1. Usually I see questions like this and it's 4...16...25, etc.

  • 4
    $\begingroup$ Welcome to Puzzling.SE! Could you provide proper attribution so we know where this puzzle came from? $\endgroup$
    – Riley
    Commented Jun 3, 2018 at 14:12
  • $\begingroup$ It appears to be from matrix67.com. $\endgroup$ Commented Jun 3, 2018 at 18:11
  • 3
    $\begingroup$ The test seems to be the copyright of Nathan Haselbauer and Mike Dickheiser, and in the introduction to the test they write "We ask that you do not share your answers in any public forum in order to maintain the integrity of the test." $\endgroup$ Commented Jun 3, 2018 at 21:42
  • $\begingroup$ Changing the question does not absolve you of the need to provide attribution for content you're posting here that is not your own. Yes, you need the permission of the original content creator to post their work here. But even if you have it (and merely being unaware that you don't have it is by no means the same thing!), you still need to disclose the original source. $\endgroup$
    – Rubio
    Commented Jun 4, 2018 at 0:56
  • $\begingroup$ I'm putting this question on hold until proper attribution of its original source is provided. — It looks like you're asking us to solve a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. Please edit to include (at minimum) where this came from—any additional context you can provide is also helpful to solvers—then vote to reopen. Posts which use someone else's content without attribution are generally deleted. $\endgroup$
    – Rubio
    Commented Jun 4, 2018 at 0:57

1 Answer 1


I believe the answer is



$\sqrt{3968 + 1} = 63$
$\sqrt{63 + 1} = 8$
$\sqrt{8 + 1} = 3$
$\sqrt{3 + 1} = 2$

P.s: I guess this is a new question not related to matrix67.com

  • 1
    $\begingroup$ Yeah, I changed the question once I was informed the authors didn't want the answers publicly posted. $\endgroup$ Commented Jun 4, 2018 at 0:06

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