imadummy wrote:From a Veritas Prep workbook:

For how many integers, n, does 2^n = n^2

A. Zero
B. One
C. Two
D. Four
E. Eight

yeah, this problem is somewhat unappetizing, as trial and error is the only way to solve it.

this problem strikes me as something unlikely to show up on the gmat, because of its combination of the following 2 factors:
1, soluble only by trial and error;
2, technically an infinity of trials to do (although it becomes perfectly clear after trying x = 3, 4, 5 that no bigger numbers will work).

1) this is not to say that the gmat will never feature a trial-and-error problem, because of course it often does. however, if there is a problem based on trial and error, then it will almost certainly feature a finite set of 'trials', usually based on some sort of combinatorial issue. here is an example from another forum (i don't normally like to cross-post, but i don't recall any equally juicy examples i've posted here).

2) and of course there are problems that would feature an infinity of trials were they to be solved by trial and error - but these problems are usually soluble with some sort of theory / number properties / something else that obviates the need for trial-and-error in the first place.

incidentally, it is possible to use number properties to figure out that 'n' has to be a power of 2.

here's how:
consider prime factorizations.
obviously, the prime factorization of 2^n contains only 2's, and no other primes.
but because 2^n and n^2 are equal, this means that the prime factorization of n^2 also contains nothing but 2's, which means that n itself must be a power of 2.

therefore, the only integers you have to try are 1, 2, 4, 8, ...
which makes your life a lot easier.

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this insight makes the problem seem a bit more gmat-like.