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Got this as a question in an IQ test and cannot figure it out. Does anyone know?

12, 26, 31, 76, 77, 94, 101.

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    $\begingroup$ The question in the test only asked whether there that sequence follows any logic, so a "yes", "no" question? And how much time does someone have to answer this question? Because I am confident that people can come up with different patterns, which would fits this, given enough time. $\endgroup$ – fibonatic Mar 4 '15 at 2:58
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    $\begingroup$ math.stackexchange.com/questions/1174409/… $\endgroup$ – Gamow Mar 4 '15 at 9:18
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    $\begingroup$ @Gamow In comments to that question, the OP was directed to ask it here. $\endgroup$ – KSmarts Mar 4 '15 at 19:32
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    $\begingroup$ Is this a question from an on-going contest? $\endgroup$ – Joel Reyes Noche Mar 5 '15 at 2:19
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    $\begingroup$ If no was given as an option (yes/no question), I'd go with no, since I've spent about 20 minutes on it. Also, no else seems to have figured it out. $\endgroup$ – ghosts_in_the_code Mar 12 '15 at 14:56
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No there is no logic to this sequence, You might try some overly complicated algorithm to figure it out but that's not the point. The point of an IQ test is to test someone's IQ and for that they don't tend to ask for over complicated mathematical algorithms.

To be thorough:

You can try to find the difference between the numbers which are 14, 5, 35, 1, 17, 7. which have nothing logical about them.

you can try to add the digit (12 = 1+2 = 3) which would be 3, 8, 4, 13, 14, 13, 2 (or might be 11 (10 + 1)).

They are not part of any mathematical sequence such a Fibonacci, primes or any other sequence i know. Therefor I conclude the answer is No there is no logic.

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  • $\begingroup$ I really doubt this is the answer. I think the question should really be asked "what is the logic to this sequence?" $\endgroup$ – Rand al'Thor Mar 13 '15 at 11:02
  • $\begingroup$ That is not the question, I've tried to answer the question asked. Granted it might be a way of interpreting it but it's not what is asked. $\endgroup$ – Vincent Mar 13 '15 at 11:11
  • $\begingroup$ True, and for that reason I've upvoted, but I still don't think it's what the OP is looking for. $\endgroup$ – Rand al'Thor Mar 13 '15 at 11:18
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If you consider an equation to be logical enough, then the following may explain the sequence.

If $a_n$ is the $n$-th number in the sequence, then

$a_n=\lfloor{\left(1.42n+0.37\left(-1\right)^n\right)\left(11+n\right)-0.2521n^{2.812}-2.3p_{(n+k)}+0.5}\rfloor$

where $p_{(x)}$ is the $x$-th digit of $\pi$, and $k=5388843$.

For example, if $n=3$, then

$a_3=\lfloor{\left(4.26-0.37\right)\left(14\right)-0.2521\left(3^{2.812}\right)-2.3p_{(3+5388843)}+0.5}\rfloor$

Since the 5388846th digit of $\pi$ is $8$, then

$a_3=\lfloor{\left(3.89\right)\left(14\right)-0.2521\left(3^{2.812}\right)-2.3(8)+0.5}\rfloor$

$=\lfloor 54.46-0.2521\left(3^{2.812}\right)-17.9 \rfloor$

$=\lfloor 31.023468\rfloor=31$, which is the 3rd number in the sequence.

Note that $k$ can have other values, such as 6049868, 11551553, or 12701077, and the same sequence will be produced (only possibly differing in the 8th and succeeding terms).

EDIT:

Use this to search for digits of $\pi$.

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    $\begingroup$ I did some brute-force simulations in Excel. $\endgroup$ – qzx Mar 13 '15 at 8:01
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    $\begingroup$ Using enough force, you can find an equation for every sequence. How/Why would someone come up with THAT equation in the first place? $\endgroup$ – Alexander Mar 13 '15 at 8:18
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    $\begingroup$ @Alexander That logical pattern came to my mind as soon as I read the question, but qzx beat me to it :-) $\endgroup$ – Daniel Daranas Mar 13 '15 at 9:03
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    $\begingroup$ Since this sequence was in an IQ test, if this is really the answer, then I guess that the IQ test probably was designed to test the IQ of alien beings from a race much more intelligent than those stupid and naive small-brained primitive animals called "humans". $\endgroup$ – Victor Stafusa Mar 13 '15 at 10:32
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    $\begingroup$ Excellent answer. As always, Puzzling.SE delivers the goods. :) $\endgroup$ – A E Mar 13 '15 at 10:43

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