Bunuel wrote:
Is \(n\) an even number?
(1) \(n^2 = n\)
(2) \(n^3 = n\)
Another approach: This is the sort of problem that can get you into trouble if you follow algebra blindly, without thinking about how numbers may enter into the equation. Of course, the solution proposed by
Bunuel is algebraic and sound, but I imagine some people, rather than think to isolate the variable by moving the unknown
to one side to solve, would look to cut corners and divide both sides by
n instead. In the former case, the statement would yield
n = 1; in the latter,
n^2 = 1. It would then appear as if choice (D) were correct, since either 1 or -1, both odd numbers, would represent valid answers to the altered Statement (2). However, a simple check with
actual numbers can allow you to see through the trap with confidence. Start by asking yourself when a number times itself
could equal itself. That is not true of any fraction or negative number, but it is true of both 0 and 1. With one valid solution being even and the other odd, Statement (1) is clearly NOT sufficient. Such reasoning not only helps to eliminate (A) and (D), but it can also be applied readily to Statement (2), same numbers and all. That is, both
\(0^3 = 0\) and \(1^3 = 1\).
We could even toss in -1 on this one, since \(-1^3 = -1\), but in terms of odds or evens, it makes no difference. The picture is just as murky as before. We can thus eliminate choice (B).
Between (C) or (E),
look to argue against one or the other being true. That is, we have established that either condition allows for solutions of 0 or 1, so by putting the two statements together, can we whittle the answer pool down to one value or the other to answer the question? Of course, we cannot in this case: 0 or 1, even or odd, will still hold true. Reminder:
do not overlook the obvious when approaching Quant questions. Sometimes an intuitive, numbers-based approach can allow you to solve the question more quickly and confidently than what you may perceive to be the
correct approach.