Find out the next number in the following Series

5 , 7 , 12 , 13 , 9 , ? .


Square root.


closed as too broad by Dan Russell, Will, GentlePurpleRain, Engineer Toast, Deusovi Jul 26 '16 at 15:54

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  • $\begingroup$ note that this is tagged with lateral thinking, which means the answer might not be purely algebraic. I have been trying to find patterns in the english words and roman numerals but currently I have nothing. $\endgroup$ – as4s4hetic Jul 26 '16 at 12:35
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    $\begingroup$ I voted to close because as too broad because there are already at least two answers that appear to be valid. Please refine or extend the question. $\endgroup$ – Dan Russell Jul 26 '16 at 14:07
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    $\begingroup$ @Jishnu Chandran: If you want to add a hint or otherwise clarify your question, log in to do it.  We just rejected a suggested edit from an anonymous user.  You can freely edit your own posts, but, for your protection, this must be done under the original user account.  It looks like you may have tried to edit your question without logging into your account. $\endgroup$ – Peregrine Rook Jul 26 '16 at 17:41
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    $\begingroup$ @KNeerajLal : I have updated one hint, this one can clearly solve the answer. $\endgroup$ – Jishnu Chandran Jul 27 '16 at 6:13
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    $\begingroup$ @JishnuChandran Yes, I get it now :) Please reopen the question mods. :) $\endgroup$ – K Neeraj Lal Jul 27 '16 at 8:00



Let's name the numbers

$x_1, x_2, x_3, x_4, x_5, x_6$.
We need to find $x_6$.

Adding the numbers 2 by 2 in such a a way that their indexes sum up to 7 we get:
$x_3 + x_4 = 12+13 = 25 = 5^2$
$x_2 + x_5 = 7+9 = 16 = 4^2$
$x_1 + x_6 = 5+4 = 9 = 3^2$
It's about perfect squares

  • $\begingroup$ Number order is a bit strange if this is th answer $\endgroup$ – Fabich Jul 26 '16 at 12:20
  • $\begingroup$ I admit, is kind of a long shot, but it's all I have. $\endgroup$ – Marius Jul 26 '16 at 12:21
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    $\begingroup$ i love this answer. if it's not the right one, it should be. $\endgroup$ – Anthony Jul 26 '16 at 12:22
  • $\begingroup$ I agree @Anthony,but the outcome is not correct. $\endgroup$ – Jishnu Chandran Jul 26 '16 at 12:32
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    $\begingroup$ @Marius and fine effort in your perspective.Should i provide a hint. $\endgroup$ – Jishnu Chandran Jul 26 '16 at 12:33



5 + 7 = 12, 13 + 9 = 22


If We want go on with sequence of numbers, I think It goes like that:

5 + 7 = 12 , 13 + 9 = 22, 23 + 11=34, 35 + 15 = 50
If 5,7=12 and the 13,9 I think You did 7+2, 12 -> 2. So, 23+11 , 9 (13+9) and 2 (22 -> 2).

  • $\begingroup$ Welcome to Stack Exchange, and Puzzling Stack Exchange in particular.   (The first part of) your answer is no less plausible than any of the others that have been posted.  You’ll find that the questions here are usually trickier than that, but not always; there’s no harm in offering a plausible guess if the correct answer hasn’t been revealed yet.  But the second part of your answer (the “Other” edit) is very confusing.  You seem to be using , (comma) to mean + (plus) and - (hyphen) to delimit a list,  … (Cont’d) $\endgroup$ – Peregrine Rook Jul 26 '16 at 18:57
  • $\begingroup$ (Cont’d) …  except where you say “…22, 23…” and “…34, 35…”, in which you seem to be using , (comma) to delimit a list.  And you do use + after that, so why are you using , (comma) to mean + (plus) in the first place? If you’re extrapolating from $n_4=n_3+1$ to assume $n_7=n_6+1$ and $n_{10}=n_9+1$, you should say so. And where are you getting your suggested values for $n_8$ and $n_{11}$? Your last line (starting with “If”) doesn’t make any sense to me at all.  You probably shouldn’t be using backticks around “5,7=12 and the 13,9” like that. What does “2” have to do with anything? $\endgroup$ – Peregrine Rook Jul 26 '16 at 18:58
  • $\begingroup$ P.S. I see that you’re just learning English. For your information, in ordinary English sentences, we capitalize the first word, (most) names (e.g., “Stack Exchange”), many abbreviations (e.g., “Q&A” for “question and answer”), and the pronoun “I” — and not much else — so “We”, “It”, and “You” should not be capitalized. $\endgroup$ – Peregrine Rook Jul 26 '16 at 19:00

I think 4 is the answer. 12 to 7 difference is 5 so 13 to 9 difference is 4.

  • $\begingroup$ No.This is not the answer. $\endgroup$ – Jishnu Chandran Jul 26 '16 at 12:17
  • $\begingroup$ Oh Okay. let me think :) $\endgroup$ – Vivek Keviv Jul 26 '16 at 12:18
  • $\begingroup$ While I’ve seen some very weird questions (puzzles) and some very weird answers (some of which have been marked as correct) on this site, it’s conventional for sequences/series of numbers to work sequentially. Your answer would make a little bit of sense if $n_1-n_2=n_3$, but that’s not the case. If you’re going to suggest a non-sequential answer to a series question, you should at least acknowledge that it is a non-sequential answer and, ideally, provide a rationale as to why it is likely to be the correct answer. $\endgroup$ – Peregrine Rook Jul 26 '16 at 17:51

I just want to try :)

Is it



13 - 12 = 1; 9 - 7 = 2; x - 5 = should be 3

so x = 8.

  • $\begingroup$ Please someone find the answer I am really tired to checking if answered at any time whatsoever ... $\endgroup$ – Hilal Jul 27 '16 at 10:31




if we translate each number to a letter corresponding to its position in the alphabet, we get E G L M I and so of course we're missing N (=14) because then we can spell the word MINGLE, mingled.

  • $\begingroup$ That word is somewhat appropriate to the inputs of this puzzle. But, aside from that, you could also use gimlet, midleg, and a few non-standard ones (e.g., gimble, leming, and milage). $\endgroup$ – Peregrine Rook Jul 26 '16 at 21:43

Is the Answer 22?


12- 7 = 5 & 22 - 13 = 9

  • $\begingroup$ Your answer is mathematically equivalent to Ronronner’s answer, which was posted before yours.  That’s not a big deal in this case, the time delta was 18 seconds, and, IMO, anything under 42 seconds is a tie.  However, that post presents the result in a straightforward, constructive manner (here’s how you get from the given numbers to the missing number) while your uses a backwards approach (Poof!  Here’s the answer, which I pulled out of thin air; now, here’s an argument that suggests that it fits the question.) $\endgroup$ – Peregrine Rook Jul 26 '16 at 19:01

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