by RonPurewal Tue Jul 29, 2008 5:09 am
yep, number plugging is definitely the method with the least hassle for problems like this one.
if you like theory, you can designate an arbitrary even integer as "2k" where k is any integer.
in that case, choice b = (6k/2) + 1 = 3k + 1, which is odd if k is even, but is even if k is odd.
still, it's much easier to plug in numbers; you won't have to go far (no farther than 4) before you're left with a unique answer. that will take way less time than grinding "2k" through all the answer choices AND working out the parity (even/odd) of the resulting expressions.