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

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

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04 Mar 2012, 21:11
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If n and k are integers and (-2)n^5 > 0, is k^37 < 0?

(1) (nk)^z > 0, where z is an integer that is not divisible by two
(2) k < n
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Re: If n and k are integers and (-2)n^5 > 0, is k^37 < 0?  [#permalink]

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04 Mar 2012, 23:56
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If n and k are integers and (-2)n^5 > 0, is k^37 < 0?

First of all: $$(-2)*n^5>0$$ --> reduce by negative -2 and flip the sign: $$n^5<0$$ --> $$n<0$$;
Next: "is $$k^{37}<0$$" basically means whether $$k<0$$.

So the question becomes: If n and k are integers and $$n<0$$ is $$k<0$$?

(1) (nk)^z > 0, where z is an integer that is not divisible by two --> $$(nk)^{odd} > 0$$ --> $$nk> 0$$ --> $$n$$ and $$k$$ have the same sign and since $$n<0$$ then $$k<0$$. Sufficient.

(2) k < n --> since given that $$n<0$$ then $$k<n<0$$. Sufficient.

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Re: If n and k are integers and (-2)n^5 > 0, is k^37 < 0?  [#permalink]

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22 Jun 2012, 13:54
Bunuel I find this question on the net.......I would like to know if you think that is a good question and if it is elaborate in a good manner.

It is important to practice.

Thanks
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Re: If n and k are integers and (-2)n^5 > 0, is k^37 < 0?  [#permalink]

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23 Jun 2012, 02:58
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carcass wrote:
Bunuel I find this question on the net.......I would like to know if you think that is a good question and if it is elaborate in a good manner.

It is important to practice.

Thanks

Though the question is testing basic concepts usual for the GMAT, I still wouldn't call that question very elegant.
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Re: If n and k are integers and (-2)n^5 > 0, is k^37 < 0?  [#permalink]

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11 Mar 2013, 17:37
What about the possibility that Z is equal to 0. In that case 1) would be insufficient.
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Re: If n and k are integers and (-2)n^5 > 0, is k^37 < 0?  [#permalink]

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11 Mar 2013, 20:47
thekid1998 wrote:
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.
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Re: If n and k are integers and (-2)n^5 > 0, is k^37 < 0?  [#permalink]

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12 Mar 2013, 06:26
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 .
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If n and k are integers and (-2)n5 > 0, is k37 < 0? (nk)z > 0, where z  [#permalink]

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11 Jan 2019, 23:22
1
UB001 wrote:
If n and k are integers and (-2)n^5 > 0, is k^37 < 0?

1. (nk)^z > 0, where z is an integer that is not divisible by two
2. k < n

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
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Re: If n and k are integers and (-2)n5 > 0, is k37 < 0? (nk)z > 0, where z  [#permalink]

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11 Jan 2019, 23:29
1
UB001 wrote:
If n and k are integers and (-2)n^5 > 0, is k^37 < 0?

1. (nk)^z > 0, where z is an integer that is not divisible by two
2. k < n

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

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Re: If n and k are integers and (-2)n^5 > 0, is k^37 < 0?  [#permalink]

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12 Jan 2019, 02:27
UB001 wrote:
If n and k are integers and (-2)n^5 > 0, is k^37 < 0?

1. (nk)^z > 0, where z is an integer that is not divisible by two
2. k < n

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Merging topics.
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Re: If n and k are integers and (-2)n^5 > 0, is k^37 < 0? &nbs [#permalink] 12 Jan 2019, 02:27
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