If n and k are integers and (-2)n^5 > 0, is k^37 < 0?

Though the question is testing basic concepts usual for the GMAT, I still wouldn't call that question very elegant.

What about the possibility that Z is equal to 0. In that case 1) would be insufficient.

Z can't be zero as it is mentioned that z is not divisible by 2.

Another great DS problem that helps pinpoint the need to stay alert at the sign of the base of the exponent and whether the power itself is odd or even.

Let's begin by understanding the problem.

We have \(-2*(n)^5 > 0\). Since the power is odd and the end result is positive despite multiplying the expression \(n^5\) by (-2), then this states that n is negative.
So we want to see whether \(k^(37)<0\), which is translated to : is k negative ? since 37 is an odd number !

Statement 1 : \((nk)^z > 0\), where z is an integer that is not divisible by two

Since z is odd and the expression is positive, we can assume that the base, nk, is positive. And seeing as n is negative, we can say that k is also negative. And since 37 is an odd power, then \(k^(37)\) is negative. So statement 1 is sufficient.

Statement 2 : \(k < n\)

Since n is negative, it naturally follows that k is also negative. And since 37 is an odd power, then \(k^(37)\) is negative. So statement 2 is sufficient.

The correct answer is, therefore, D !

Hope that helped .

Analysing question stem
(-2)n^5 > 0...means 'n' is negative because only then the product of -2 and n will be positive integer...and the sign of integer raised to odd power will be same as sign of the integer.

So n < 0 .....(A)

Find:
k^37 < 0?
That is, Is k < 0?

Statement 1:
(nk)^z > 0 ...and z is odd.
That is... Product of negative "n" and "k" raised to odd integer "z" is positive.
This is possible only when k is negative, because only then the product n*k will pe positive...and (n*k)^odd integer will be greater than 0.
Hence, sufficient.

Statement 2:
k < n
Since 'n is less than 0 (from question stem analysis), k is also less than 0.
Hence k^37 < 0
Sufficient

Ans: D

So, before solving the question, lets brain storm over it a bit

If n and k are integers and (-2)n^5 > 0
That will imply that n will be -ive
(-2) * (-I)^5 will only give something > 0

Statement (1)
(nk)^z > 0, where z is an integer that is not divisible by two
z will be 1,3,5,7
here k will be - ive, (-n* -k)^Odd Integer, will give something > 0

When value of k^37 < 0, is put back in the question, it will be sufficient.

Statement (2)
k < n, Since n is -ive
k will be -ive

(-ive k)^ 37 < 0, will always be less than 0, it will be sufficient


Answer D