As Matt says, this won't be relevant to GMAT test takers, so no need to worry about any of this if you're preparing for the test, but I find it fun to sometimes talk about math problems that are beyond the scope of the test:
Matt@VeritasPrep wrote:
I actually took the first interpretation: they do mean that each key is a skeleton key that works in any of the five locks. From there, I took the minimum as the fewest (cleverly chosen) trials we'd need to run to prove this to a skeptical observer and the maximum as the worst (= longest) scenario of all the scenarios in which we pick x random and distinct arrangements and are forced to run all x to see that the keys are all skeleton.
(That wasn't well put, but you get the idea: we run x random and distinct arrangements, with x not determined beforehand, and only after running exactly x random and distinct arrangements do we see that the keys are skeleton, with the maximum being the greatest possible value of x. In other words, if the answer is, say, 15, then any 15 distinct random arrangements MUST force the conclusion that the keys are skeleton, and there is some set of 14 distinct random arrangements that does NOT force that conclusion.)
Taken that way, I thought it was a neat question, but probably one too difficult and time consuming for the GMAT. (Just the minimum seems fine to ask, though.)
That is an interesting question, though it does have a fast solution, if I'm understanding it correctly: we have 5 keys, and 5 locks, and we want to prove that each key opens every single lock. We're going to randomly try five keys in five locks in each trial, and once we've done a trial, we will never repeat it. If we keep doing trials, what is the maximum number of trials we might need to do before we have tried every key in ever lock?
The answer to that question should be 97. There are 5! = 120 ways to match the keys to the locks in general. In 1/5 of those ways, we match the first key to the first lock. So in 4/5 of those ways, or 96 ways, we do not try the first key in the first lock. We need to do that to prove our keys are skeleton keys, so if by bad luck we don't do that in the first 96 trials, we will surely do it on the 97th. But if we do 97 trials, every key must have been tried in every lock at some point, essentially by the pigeonhole principle (once we've tested more than 4/5 of possible arrangements, everything must have been tried everywhere at some point).
David (sorry to 'demolish' the problem
) - you're probably familiar with this already, but that GMATPrep problem is a simple example of a class of combinatorics problems known as 'derangement' problems, where you're interested in how many ways you can rearrange, say, the letters in ABCD where no letter remains in its original place. These get complicated quickly - with even just five letters, it would be well beyond GMAT-level math to ask a derangement question, so the GMAT version of the question, with only 4 letters, is really the only version you could ever see on the test (with fewer letters the question isn't that interesting). The question in the OP above isn't really a derangement problem, but it may have been inspired by one.
