Hello, you may be familiar with this problem from the Making Deductions III video.
I found that if you split the groups into 5-5-6 to start, and the 5 vs. 5 groups are uneven in weights, then you can split the lighter group of 5 into 2-2-1. In this case, if the 2 vs. 2 groups are even, then the remaining 1 is our light block, which means best case scenario we only need two weighings. However, in the worst case scenario we will still need 3 weighings.
Is the answer QA=QB because we care about the “limiting reactant” here, which is the worst case scenario of three weighings? And therefore that best case scenario of two weighings does not matter? Thanks!
Well your “best” case scenario isn’t really best, right? You could’ve just picked two blocks from the group of 16 and placed one on each side. If you’re lucky, you end up identifying the lighter block with just 1 weighing.
Rather than calling it the “worst case” (which is technically 8 weighings), it’s more accurate to say something along the lines of “worst case for the optimal strategy.” Anyhow, I think you have the right idea.
