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The following problem was given to me by one of my friend:

Find the wrong number in the given series:
$$6,\, 12,\, 21,\, 36,\, 56,\, 81 $$ ?
A) $21\,$ B) $12\,$ C) $36\,$ D) $56$

and told me that its answer is B) $12$

but according to me answer should be D) $56$ ; as all the other numbers in the series is exactly divisible by $3$

So, is my answer right ? or there is more reasonable approach for option B) $12$

Any hints/suggestions or help please...

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    $\begingroup$ Since it is a series, then the next number is most likely to be derived relative to the previous term more than its own individual properties. That's usually the case. Nice puzzle! :) $\endgroup$ – Mr Pie Jul 14 at 4:15
  • $\begingroup$ Odd-number-out problems present numbers as a series even if they don't use the word "series", and even if the answer is odd one out because of its own individual properties, so I don't see the word "series" as evidence that that's not the case here. It's ambiguous. Both athin's and JKHA's ideas are valid. To make this problem's issue easier to see: Q: Which is the wrong number in this sequence: 3, 6, 9, 15, 21, 28, 36, 45. rot13(A: 28: the rest are multiples of 3. B: 9: the rest are triangle numbers.) $\endgroup$ – Rosie F Jul 14 at 17:29
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The series should be:

$6, 11, 21, 36, 56, 81$

Because:

The differences will be $5, 10, 15, 20, 25$.

So the answer is:

$B)~12$

Note that:

This is given as a series, not a set, therefore it's wise to treat it as it should be.

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My guess is:

A reasonable approach, following your solution is that all those numbers minus $1$:
$5$, $20$, $35$, $55$, $80$
are divisible by $5$.
$12 - 1 = 11$, isn't.

Nevertheless, I don't know where this puzzle comes from but that won't be surprising you are right and your friend may have made a mistake... Unless PSE community finds a simplier solution :)

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    $\begingroup$ What happened to Occam's Razor? $\endgroup$ – M Oehm Jul 13 at 11:12
  • $\begingroup$ @MOehm haha, good point, that's the simpliest I've found for now, but I'm still searching :) Editing my question to take it in account ;) $\endgroup$ – JKHA Jul 13 at 11:13
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Whenever I see a question like this, I imagine there is unlimited formulas to cover each optional (randomly) selection of those numbers in any direction, and if there isn't any other limiting rules, every answer to that question is correct. take a look at this

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    $\begingroup$ Welcome to PSE! That's a good point. Perhaps to filter the illimited number of formulas for those kind of enigma, we'll select the easiest one :) $\endgroup$ – JKHA Jul 14 at 9:16
  • $\begingroup$ @JKHA You are right, the thing I wanted to bold is that, some of questions of this type, introduce multiple "easiest" answers and therefore they introduce doubt in absolute maths problem. each type of person can come to different answers. I see a lot of IQ test applications and math puzzles do not consider this, and each method of thinking can defend different correct answers $\endgroup$ – FarhadGh Jul 14 at 9:27
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    $\begingroup$ That's exact. That's good StackExchange allows a user to vote for several different answers then :) $\endgroup$ – JKHA Jul 14 at 9:35
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    $\begingroup$ It's true that you can fit polynomials to pretty much any pattern, with enough work. But while such would be brute force solutions to a math problem, they're never going to be the intended solution to a puzzle, so we summarily dismiss answers of this type; they're technically answers to the question, but not solutions to the puzzle, and there is a difference. $\endgroup$ – Rubio Jul 14 at 11:25

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