very boring, and I dont see any chances of solving it on the real exam.
you have to just follow the definition to obtain every possible set.
this gives you exactly 16 elements, and constructing further does not increase the number. this is plain from a diagram (plot a "big" set M, and two intersecting proper subsets of it, namely, A and B. S_2 already gives you every possible intersection. diagram also could be useful to build the S_2 itself. however, this solution is extremely time-consuming and anxiety on the test will almost certainly prevent you from succeeding. also it is easy to forget including empty set and ending up with answer '15', which is among the choices.
I'd be happy to look at other solutions.
EugeneKudashev wrote:power set (set of all possible subsets) is good here when you already know the answer. unfortunately, on step 1 (having S_1) you do not even have these four distinct parts. so you can't say for sure what you will get. and you can't say for sure that the collection will be finite.
volodja wrote:Yeah, problem 2 isn't hard if there is a computer at hand... But how to figure out the answer in a fast way? As to problem 1, I think it is very easy but I really don't understand it; perhaps some well-known properties of a mirror are expected to be applied or something like that.
Users browsing this forum: No registered users and 10 guests