If \(n\) is an integer, is \(n\) even?
(1) \(n^2 – 1\) is an odd integer.
(2) \(3n + 4\) is an even integer.
Because we’re only being asked to identify the parity (eveness or oddness) of a single variable, and not an expression, there’s no good rephrase. So, let’s move on to the clues.
There are two ways to approach these kinds of problems. We can take an analytical approach or we can test numbers. I’ll present both.
THE ANALYTICAL APPROACH: We’ll start with clue (1) which can be rewritten
\(n^2 – 1 = (n – 1)(n + 1) = odd\)
Because \(n\) is an integer \(n – 1\) and \(n + 1\) must be integers, and the only way to get an odd product from two integers is for both the factors to be odd. In other words, the only way this happens is \({odd}\times{odd} = odd\). Therefore, we know that both \(n – 1\) and \(n + 1\) are odd. Consequently \(n\) must be even. Clue (1) is sufficient. B, C, and E are out.
Clue (2) is a bit easier:
\(3n + 4 = even\)
\(3n = even – 4\)
An even plus (or minus!) an even is an even, so now we have
\(3n = even\)
Since 3 is odd, we have
\({odd}\times{n} = even\)
Therefore, \(n\) is even.
(2) is sufficient as well. The correct answer is D.
TESTING NUMBERS: We need to test an odd and an even. Let’s keep it simple – we’ll use \( n = 2\) for our even, and \(n = 3\) for our odd.
Clue (1):
\(2^2 – 1 = 4 – 1 = 3 = odd\)
That works – i.e. if \(n = 2\) the condition of the clue is fulfilled.
\(3^2 – 1 = 9 – 1 = 8 = even\)
That does not work – i.e. if \(n = 3\) the condition of the clue is not fulfilled. Therefore, \(n\) can not be odd. It must be even. Clue (1) is sufficient. B, C, and E are out..
Clue (2):
\(3(2) + 4 = 6 + 4 = 10 = even\)
That works – i.e. if \(n = 2\) the condition of the clue is fulfilled.
\(3(3) + 4 = 9 + 4 = 13 = odd\)
That does not work – i.e. if \(n = 3\) the condition of the clue is not fulfilled. Therefore, \(n\) can not be odd. It must be even. Clue (2) is sufficient. The correct answer is D.
Many students are confused about how many cases (numbers) they should test. Evens and odds are very well behaved (products and sums follow hard and fast rules), so, even in more complicated problems, one even number and one odd number are enough to determine sufficiency. Of course, there are exceptions. This is why I prefer the analytical method. It’s a bit more complex, and it may seem tough and long winded now, but it will pay dividends later. Furthermore, rephrasing will only payoff if you are fluent in the analytical method.