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In an alternative universe where the following is true:

1 # 1 = 3

2 # 2 = 7

1 # 2 = 5

What would ? be?

5 # 5 = ?

YOU DO NOT NEED THE FOLLOWING TO FINISH.

Extra equations:

3 # 3 = 11, 4 # 4 = 15

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    $\begingroup$ I would -1 this if I had the rep because I hate "logic" puzzles that gratuitously use symbols like + to obfuscate the fact that they're asking you to find an unknown operator. $\endgroup$ – R.. GitHub STOP HELPING ICE Oct 11 '14 at 1:23
  • $\begingroup$ @R What else was I supposed to use? Was I supposed to use #? $\endgroup$ – warspyking Oct 11 '14 at 1:47
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    $\begingroup$ It could be absolutely anything: it's clear that one or more of the symbols doesn't have the meaning we're familiar with, but you haven't given us any reason to suppose that any of the symbols does. $\endgroup$ – Peter Taylor Nov 9 '14 at 8:37
  • $\begingroup$ I happened to notice this question listed in my top answers and clicked to see what it was out of curiosity. I noticed you got lots of down votes for the sort of question many ask and which I see no issue in. It makes me wonder if people are responding to my facetious answer by seeing it as a flaw in your question worthy of downvoting, which is unfair since this is a common sort of puzzle question and, while a little easy, is defiantly worth thanking someone for creating. So in case you did somehow get penalized for my good nature ribbing let me say I'm sorry :). $\endgroup$ – dsollen May 9 '17 at 14:52
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  1. Seems to follow the formula: $(x+y) + (x+y-1)$

By the way, as an alternative let's say your plus is $+^{\prime}$, now we can define the operator: $x+^{\prime}y = 2(x+y)-1$

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  • $\begingroup$ 19 is indeed correct, although I'd need to test the forumula, it seems to work with single digit versions, could you solve 10 + 10 = ? $\endgroup$ – warspyking Oct 10 '14 at 21:17
  • $\begingroup$ it would be 39 according to this $\endgroup$ – d'alar'cop Oct 10 '14 at 21:19
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    $\begingroup$ Yep your correct. For a visual demonstration, you'd actually add them, and then "reproduce" for example, 1 + 1 is 2, so then 2 can make another 1, so now there's no way to reproduce with the new 1, therefore 2 + 1 is 3, making it 1 + 1 = 3 $\endgroup$ – warspyking Oct 10 '14 at 21:35
  • $\begingroup$ I actually wrote a program in Lua, with this operator, I called it +rep+ where rep mean "replicate" or "reproduce" $\endgroup$ – warspyking Oct 11 '14 at 1:58
  • $\begingroup$ @warspyking Interesting, do you think you could explain the "reproduce"/"replicate" again? $\endgroup$ – d'alar'cop Oct 11 '14 at 2:12
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I think the answer is clearly 17.

My Logic is that for each number digit you replace it with the next number in the fibinaci sequence, add as usual, then add by one. so we get 

1+1+1=3 (second number in fibonacci is 1)
3+3+1=7
1+3+1=5

and thus 8+8+1=17

Or maybe the answer is 23! because the formula is $(x+y) + (x+y-1)$, but of course in this universe were uses the octal numbering system!

Or maybe they use a base 10 numbering system, but they order their numbers from 0-9 like this:

8,3,1,2,0,5,7,6,9,4

In which case the answer is 34!

All are correct given the 3 equations which are the 'only ones I need' to solve the puzzle :)

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    $\begingroup$ that's quite clever :) $\endgroup$ – d'alar'cop Oct 10 '14 at 22:07
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    $\begingroup$ Interesting idea. Just goes to show you can justify just about any answer you want on a sequence like this. $\endgroup$ – Kevin Oct 10 '14 at 23:00
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    $\begingroup$ If I could mark this accepted too, I would! $\endgroup$ – warspyking Oct 11 '14 at 1:38
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Without looking at the "extra equations", I have come up with this solution:

Evaluate the sum and look for the corresponding prime number.

The first few prime numbers are 2 3 5 7 11 13 17 19 23 29.

1 + 1 = 2nd prime number = 3

2 + 2 = 4th prime number = 7

2 + 1 = 3rd prime number = 5

Therefore 5 + 5 = 10th prime number = 29

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