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Two of either is one, one of each is the other.

To what do I refer?

Note: There are two different solutions.

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closed as off-topic by xnor, Engineer Toast, Mark N, Raystafarian, mmking May 28 '15 at 13:48

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Okay, two possible solutions:

Solution 1:

0 and 0. Two of either (0+0) is one of the 0s, one of each (0+0) is the other 0! Kind of cheating, I know.

Better Solution 2:

1 and -1. Two of either (1*1 or -1*-1) is one (1), one of each (-1*1) is the other (-1).

Possible Solution 3 (thought of independently of Andreas's answer =P):

True and False, split by "==". (True == True) is True, (False == False) is True, (True == False) is False.

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  • $\begingroup$ Solution 1 might actually be a third answer to the riddle, but I'll have to run across it a few times just to make sure. Solution 2 is correct, though. But that leaves another solution to be found. $\endgroup$ – Andrew Larsson May 27 '15 at 23:07
  • $\begingroup$ I like solution 1 more, since it uses addition. Two 1's isn't really 1*1, that's one 1. Just like -1 + 1 is 0 and not -1. Addition is implicit in the question. $\endgroup$ – Andreas May 27 '15 at 23:36
  • $\begingroup$ Solution 1 is actually provable mathematically! Let x = "one" and y = "the other". The question gives us three equations: 2x = x, 2y = x, y + x = y. Using the last equation: y + x = y ==> x = y - y ==> x = 0. Substitute into the second equation: 2y = 0 ==> y = 0. Double check with the first question: 2(0) = 0 ==> yes, that is true! Thus, x = 0, y = 0. $\endgroup$ – VictorHenry May 27 '15 at 23:49
  • $\begingroup$ Even though my question got put on hold, I'm marking your answer as correct due to this meta post meta.puzzling.stackexchange.com/questions/1383/… $\endgroup$ – Andrew Larsson May 28 '15 at 17:03
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Another answer:

$\frac{1}{2}$ and $\frac{1}{2}$. Two of either of these is $1$, and one of each (i.e. each one of them) is the other one of them.

I suspect the two solutions you intended were this and @VictorHenry's second solution.

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  • $\begingroup$ This solution, while seemingly correct, doesn't really meet the requirement of the second part of the riddle "one of each is the other." This has been a common answer to this riddle when asked to friends and family however. $\endgroup$ – Andrew Larsson May 27 '15 at 23:05
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If we're free to pick operators, then:

Exclusive OR: true XOR true = false, false XOR false = false; false XOR true = true

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  • $\begingroup$ Argh, I literally just now thought of this, you THIEF! $\endgroup$ – VictorHenry May 27 '15 at 23:22
  • $\begingroup$ This works just the same with the AND operator, but is effectively the same answer as "Solution 2" from @VictorHenry. $\endgroup$ – Andrew Larsson May 27 '15 at 23:22
  • $\begingroup$ It doesn't work with AND, because true&true isn't equal to false&false, which should both be equal one of the two. $\endgroup$ – Andreas May 27 '15 at 23:24
  • $\begingroup$ Hrm, actually mine was different now that I think about it. I'll add mine as an edit, but given Andrew's response to this, it's probably not right. $\endgroup$ – VictorHenry May 27 '15 at 23:35
  • $\begingroup$ Oh, yeah, didn't see that. I meant that it works like saying "false false is true" because of double-negatives. $\endgroup$ – Andrew Larsson May 28 '15 at 16:23

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