7
$\begingroup$

I've found the following puzzle:

Puzzle

I think the answer is

5.
In the first row, the 1 can be explained by the red ball.The second row is a bit different. Instead of looking at the second row to explain the number 2, look at the first row. You'll see 2 red rings. The third row is similar to what we came up with. Ignore the objects in the third row and count the red rings in row 1 and 2. We'll see 4 altogether. The fourth row is like the first row; count all the red balls you've seen from that point. We can count 5.

But I'm just not satisfied with what i have here. Maybe you guys can come up with something?

$\endgroup$
49
$\begingroup$

The answer could be

3

because you can think of the red balls and red rings as

digits in a binary representation. A ball is 1, a ring is 0.
001 is 1
010 is 2
100 is 4
011 must be 3

$\endgroup$
  • 12
    $\begingroup$ I would be shocked if it was not the answer. $\endgroup$ – oleslaw Oct 19 '16 at 10:05
  • 2
    $\begingroup$ @oleslaw I would be surprised too! $\endgroup$ – marcoresk Oct 19 '16 at 10:06
  • 9
    $\begingroup$ I'd be surprised if this isn't the intended answer too. But really, that just makes it a bad question, because there isn't nearly enough information to determine that. For example: what (aside from cultural traditions surrounding binary representations) tells you that you should add the digits instead of multiplying them to get 2? Or square them, add them and take the square root to get sqrt(5)? Without any other examples with more than one ball in the same row this can't be determined. $\endgroup$ – Nathaniel Oct 19 '16 at 11:40
  • $\begingroup$ Beat me to it... $\endgroup$ – Hosch250 Oct 19 '16 at 14:36
  • 2
    $\begingroup$ Over 39 upvotes... My goodness $\endgroup$ – bill bsl Oct 19 '16 at 17:58
15
$\begingroup$

Or it could be a "trick question" so the answer is simply 8, since the number doubles each time. The red ball and rings are just a distraction.

$\endgroup$
  • $\begingroup$ The distractions are called "red herrings". $\endgroup$ – EKons Oct 20 '16 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.