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(-2) n^5 > 0, is k^37 < 0?

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(-2) n^5 > 0, is k^37 < 0?

by sanju09 » Tue Feb 08, 2011 12:13 am
If n and k are integers and (-2) n^5 > 0, is k^37 < 0?

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

[2] k < n.



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by jaxis » Tue Feb 08, 2011 12:22 am
sanju09 wrote:If n and k are integers and (-2) n^5 > 0, is k^37 < 0?

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

[2] k < n.



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Since (-2) n^5 > 0, n^5 is is negative => n is negative.

1) Z is odd.(as it is integer and not divisible by 2)
(nk)^Z >0 =>nk is positive => K is negative -- Sifficient.

2)k<n and n is negative => k is definitely negative -- Sifficient.

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by Night reader » Tue Feb 08, 2011 11:05 am
sanju09 wrote:If n and k are integers and (-2) n^5 > 0, is k^37 < 0?

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

[2] k < n.



https://www.platinumgmat.com
given: (-2)n^5>0 only possible when n^5>0 or n>0 is k^37<0 OR k<0 (since 37 is the power of odd number and will keep sign of the base number)

st(1) (nk)^z>0, z is odd --> nk>0 only possible when k<0 Sufficient
st(2) k<n OR k<0 Sufficient

D
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