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GMAT Practice Question (One of Hundreds)
If n and k are integers and (-2)n^{5} > 0, is k^{37} < 0?
(nk)^{z} > 0, where z is an integer that is not divisible by two
k < n
Correct Answer: D
The question stem can be simplified to the following: "if n and k are integers and n < 0, is k < 0?"
It is important to begin by simplifying the question.
Since k is raised to an odd power, k^{37} will always be less than 0 if k is less than 0. Likewise, k^{37} will always be greater than 0 if k is greater than 0.
So, the question can be simplified to: is k < 0?
k^{(odd integer)} < 0 if k < 0
k^{(odd integer)} > 0 if k > 0
The question can be simplified even more. Since (-2)(negative number) > 0 and (-2)(positive number) < 0, you know n^{5} is a negative number. This means that n < 0. If n were greater than 0, the statement (-2)n^{5} > 0 would never be true.
By saying that "z is an integer that is not divisible by 2," Statement (1) is saying that z is an odd integer. So, any base raised to z will keep its sign (i.e., whether the expression is positive or negative will not change since the base is raised to an odd exponent).
z/2 = not integer if z is odd
z/2 = integer if z is even
Remember that (nk)^{z} = (n^{z})(k^{z}). So, Statement (1) says that (n^{z})(k^{z}) > 0. It is important to know that there are two ways that a product of two numbers can be greater than zero:
Case 1: (negative number)(negative number) > 0
Case 2: (positive number)(positive number) > 0
Since you know that n < 0, we are dealing with Case 1 and Statement (1) can be simplified even further:
(negative number)(k^{(odd exponent)}) > 0.
Since k will not change its sign when raised to an odd exponent, the equation can be simplified even further:
(negative number)(k) > 0. k must be a negative number. Otherwise, this inequality will not be true.
To summarize in algebra:
(nk)^{z} > 0
(nk)^{z} = (n^{z})(k^{z})
(n^{z})(k^{z}) > 0
(negative number)(negative number) > 0
or (positive number)(positive number) > 0
(negative number)(k^{(odd exponent)}) > 0
(negative number)(k) > 0
k is negative
Since k is a negative number, k^{37} < 0. Statement (1) is SUFFICIENT.
Statement (2) says that k is less than n. Since you know that n is less than 0, Statement (2) says that k is less than a negative number. Only a negative number is less than another negative number. So, k must also be a negative number. Consequently, k^{37} will always be less than 0 since (negative)^{odd} < 0. Statement (2) is SUFFICIENT.
Summarizing in algebra:
k < n < 0
Since Statement (1) alone is SUFFICIENT and Statement (2) alone is SUFFICIENT, answer D is correct.