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Given this pattern:

0, 1, 0, 2, 2, 2, 3, 4, 5, 6, 6, $n$

Find the next item at $n$

Pattern/sequence explanation required

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closed as too broad by mdc32, xnor, leoll2, Engineer Toast, Aggie Kidd May 4 '15 at 13:56

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ -1. These puzzles seem to be completely random and have multiple solutions. Even the "right" answer seems no more logical than a polynomial created to fit the data. $\endgroup$ – mdc32 May 3 '15 at 5:15
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I think the answer is

12

Because...

If you were to group all numbers in batches of 3, add them up and then get the average between them (rounding up), you get a sequence of it's own which goes 1(?) 2 3 4. Done as follows:
0 1 0 = ? (I'm assuming 1)
2 2 2 = 6 / 3 = 2
3 4 5 = 12 / 3 = 4
6 6 12 = 24 / 3 = 8

I'm probably horrible mistaken, but I thought I'd give it a shot.

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  • $\begingroup$ Correct! And (0 + 1 + 0) / 3) should be rounded to 1 (as you said). It should be noted that it was the ceiling average (all decimals round up). $\endgroup$ – Zach Gates May 3 '15 at 4:05
  • $\begingroup$ @ZachGates No way... I'm kind of amazed actually. I'm bad at puzzles haha. The reason I put one to begin with was if the averages would be a sequence, 0, 2, 4, 8 wouldn't make any sense. Which is why I assumed 1. $\endgroup$ – MisterBla May 3 '15 at 4:08
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I feel like there may be multiple solutions to this puzzle, but I'll propose one I found.

The next number is:

6

Because

Each group of three adds up to a multiple of 6, alternating between consecutive integers and the same 3 digits repeated.
- 0 + 1 + 0 = 6^0 (unsure about this)
2+2+2 = 1*6
3+4+5 = 2*6
==> 6+6+6=3*6

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  • $\begingroup$ Not quite, but you're onto something with the groups of 3! $\endgroup$ – Zach Gates May 3 '15 at 2:58

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