MHB Solve the ANUTter Puzzle: 5 Matches & 2 = 6

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The puzzle involves rearranging 5 matches to demonstrate that 2 equals 6. Participants are challenged to move only one match to achieve balance. The discussion highlights attempts and solutions, with some playful banter among users. The final solution presented is $11=\text{XI}$, showcasing the clever manipulation of the matches. The thread emphasizes creativity in problem-solving within the constraints of the puzzle.
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5 matches arranged to show 2 = 6:
|| = \/|

Move 1 match only such that both sides are equal...
 
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|| ≠ \/
 
[sp]$11=\text{XI}$[/sp]
 
Olinguito, you cheated :)

Klass, good try...but not mine...
 
$$1 = \sqrt{1}$$
 
Right on, buddy!
 

Thread 'Significant figures for inherently bound values'

I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...
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