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