Data Sufficiency

BTGmoderatorDC wrote:If a^n ≠ 0 and n is a positive integer, is n odd?

(1) a^n + a^(n+1) < 0

(2) a is an integer.

OA C

Source: Manhattan Prep

Let's take each statement one by one.

(1) a^n + a^(n+1) < 0

a^n + a^n*a < 0

a^n*(1 + a) < 0

We see that the product of a^n and (1 + a) is less than 0 or is negative.

So, either a^n is negative and (1 + a) is positive or vice-versa.

Case 1: Taking a as negative and (1 + a) as positive; Say a = -2;

a^n*(1 + a) < 0 => (-2)^n*(1 - 2) < 0 => -(-2)^n < 0.

For -(-2)^n to be negative, (-2)^n must be positive; thus, n must be even. The answer to the question: Is n odd? is No.

Case 2: Taking a as positive and (1 + a) as negative; Say a = -1/2; note that the question only specifies that n is an integer.

a^n*(1 + a) < 0 => (-1/2)^n*(1 - 1/2) < 0 => [(-1/2)^n[/2 < 0.

For [(-1/2)^n]/2 to be negative, (-1/2)^n must be negative; thus, n must be odd. The answer to the question: Is n odd? is Yes.

No unique answer. Insufficient.

(2) a is an integer.

Certainly insufficient.

(1) and (2) together

Case 2 discussed in Statement 1 is not applicable now.

We cannot take a as positive integer and (1 + a) as negative integer and maintain a^n + a^(n+1) < 0. Note that a cannot be 1, else a^n + a^(n+1) ≠ 0. Thus, a < - 1 and n is even. The answer to the question: Is n odd? is No. Unique answer. Sufficient.

The correct answer: C

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: GRE Manhattan | ACT Prep Courses San Francisco | IELTS Prep Courses Boston | Seattle IELTS Tutoring | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.