New answers tagged

0

The worst-case scenario requires 8 weighings to identify the balanced permutation, and a 9$^{th}$ weighing to verify it. As Penguino pointed out: A balanced permutation will not have each of the four balls on one side lighter than a different one on the right. Thus, 8642 cannot balance because $8>7$, $6>5$, $4>3$, and $2>1$. But 8641 could ...


4

The minimal number of needed stunts is This can be done in the following manner: This is also optimal, because: Old Answer (not taking into account the rope strength) The minimal number of needed stunts is This can be done in the following manner: This is also optimal, because:


1

No, it's not possible: If you start by weighing 3 vs 3 and it comes out equal, you only have two weighings to find the fake coin (of unknown weight) among six. In weighing 2, if you have more than three potential fakes on the scale, and it comes out unbalanced, then the remaining weighing (with only three potential results) will not allow you to pick the ...


0

I'm not entirely sure how to single out the exact ball here as the description given seems to imply that I have four groups of three balls and only three times to use the scale in order to determine the ball. Based on that, I can use standard logic to figure out which group the different ball is in. Let's say that our groups are labeled A, B, C, and D, and ...


5

Otherwise, In any event,


Top 50 recent answers are included