Question 1117550
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Arbitrarily choose one of the first two numbers as #1.  Attempt to open the lock with the first two numbers in the arbitrarily chosen order and do this once for each possible 3rd number.  If the order of the first two numbers is incorrect, this will fail 30 times.  Then reverse the order of the first two numbers and try again once for each possible 3rd number.  Worst case, this will fail 29 times and be successful on the 30th try.  So the absolute maximum number of attempts, given that he never loses his place requiring him to start over, is 59 failures followed by 1 success.  60 tries.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
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