@polylogarithmique you can actually make that a log(3) because you can split into groups of 17, 17, 16, weigh the two 17's, then split one into groups of 6, 6, 5, etc, and do it slightly faster that way.
@polylogarithmique Max would then be 17, 17, 16 -> 6, 6, 5 -> 2, 2, 2 -> 1 = 8 weighings. I don't think this is optimal but it is one way to do it.
@ChessMathNerd yes but is 2/log(3) smaller than 1/log(2)?
Pennies, lightbulbs... we know the drill by now :)
@CalbernandHowbe and @Bishop1964, congratulations on getting it right. Though you did kind of cheat Bishop, I guess I should change my questions ever so slightly so that you can't just post the video and get the answer right.
@TakeThePawnOrLose @CalbernandHowbe has an incorrect solution because there could be good pennies in the bad package. It doesn't say they are all bad.
@ChessMathNerd I thought it was kind of implied that all the pennies are bad, but I guess I should have. I apologize to anyone who disagrees with the correct answer because of that misconception, that is my fault.
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