If an ≠ 0 and n is a positive integer, is n odd?

Bunuel, can you please explain why isn't the correct ans 'A'?

Thanks.

\(a^n(1+a)< 0\) means that there can be two cases:

\(a^n > 0\) and \(1+a<0\)

OR

\(a^n < 0\) and \(1+a>0\)

Sorry forgot to copy you VeritasPrepKarishma

If an ≠ 0 and n is a positive integer, is n odd?

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

(2) a is an integer.

Dear p2bhokie,

Please find the reasoning below

(1) If A lies between -1 and 0, then the mentioned equation will be negative for all odd values of N and positive for even values of N. We'll concentrate on negative values.

Lets directly assign values to a and n

A = -0.1 N = 1
= (-.1)^1 + (-.1)^2
= -.1 + .01
= - .09

If A is lesser than -1, then the equation will be negative only for Even values of N

A = -2 N = 2
= (-2)^2 + (-2)^3
= 4 - 8
= -4

Thus (1) alone will not be able to answer whether N is odd.

Considering (2)

If A is an integer, there will be various combinations of A and N to produce a negative value

Combining 1 & 2 will eliminate the possibility of A being a number between 0 and 1.

Thus this combination will be suffice to answer.

Hope it helps you.

The solution is attached.

There are 2 variables (a and n) in the original condition. In order to match the number of variables to the number of equations, we need 2 equations. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that C is the correct answer choice. Using both the condition 1) and the condition 2), we get n=even and n+1=odd. The answer becomes no and the conditions are sufficient. Thus, the correct answer is C.

- For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Thanks a lot for this explaination.

I had a query that when a^n(1+a)<0
this means either a^n<0
or 1+a<0
hence a<-1 doesn't this eliminate the possibility of a lying in the range -1<n<0?????

Please help.

Quality Question.

Here is my solution to this one =>

Given data ->
a*n≠0-> Both are non zero numbers
n=> positive integer
a=> integer or non integer


We need to see if n is odd or not.

Statement 1->

a^n+a^n+1<0

Hence a must be negative

If a=> (-∞,-1) => n must be even so that n+1 will be odd
If a=>(-1,0)=> n must be odd so that n+1 will be even


Hence not sufficient

Statement 2->
a is an integer
No clue of n
Hence not sufficient

Combining the two statements ->
a<0 and an integer -> a can never be -1
Hence a<-1
Thus n must be even

Hence sufficient


Hence C

Hello Karishma mam,

If statement-1 changes to a^n + a^(n + 1) > 0?

Hi.
See if You change the statements to a^n + a^(n + 1) > 0 -> N can be either even or odd for a = positive integer.
But beware that a^n + a^(n + 1) > 0 does not mean that a must be positive.
a can be negative too
E.g a=-10 and n=9

The answer to your Question would be E as a can be positive or negative and n can be even/odd.

Regards
Stone Cold

(1) INSUFFICIENT: Factor an out of the left-hand side to yield an(1 + a) < 0. For this inequality to be true, the individual terms an and 1 + a must have opposite signs. Consider the two possible cases:
(i) an is positive and 1 + a is negative: If 1 + a < 0, then a < –1. Because a is negative, it follows that n must be even (since an is positive).
(ii) an is negative and 1 + a is positive: In order for an to be negative, a itself must be negative. A positive number will never turn negative when raised to a power (recall that a negative power makes a positive number smaller, but it doesn’t change a positive number to a negative one). In this case, n would have to be odd in order to make an negative.

Because n may be either even or odd, the statement is insufficient.

(2) INSUFFICIENT: This statement provides no information about n.

(1) AND (2) SUFFICIENT: Examine case (ii) of statement 1 further (the case in which a is negative and n is odd). If 1 + a > 0, then a > –1. At the same time, a itself must be negative, so -1 < a < 0.

Statement 2 specifies that a is an integer. There are no integers between –1 and 0, and so case (ii) cannot be valid. Only case (i) is possible. As a result, n must be even, so the two statements together are sufficient.

The correct answer is C.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.