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If p is an integer and m=-p+(-2)^p, is m^3>1? 1. p is

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If p is an integer and m=-p+(-2)^p, is m^3>1? 1. p is [#permalink] New post 06 Sep 2008, 17:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If p is an integer and m=-p+(-2)^p, is m^3>1?

1. p is even
2. p^3<=-1

a) 1 is sufficient
b) 2 is sufficient
c) 1 & 2 together are sufficient
d) Either 1 or 2 is sufficient
e) Neither 1 or 2 is sufficient
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Re: Math Question - Data Sufficiency [#permalink] New post 06 Sep 2008, 19:51
HVD1975 wrote:
If p is an integer and m = -p + (-2)^p, is m^3 > 1?

1. p is even
2. p^3<=-1


1: p could be - or 0 or +ve even integer.
if p = -ve, m > 1
if p = 0, m = 1
if p = +ve, m > 1
so nsf..

2: if p^3 <= -1, p is -ve and < 1. in that case, m >1.
suppose: p = -2 (the highest possible value when p is -ve).
m = -(-2) + (-2)^(-2) = 2 + 1/4 = 2.25, whcih is > 1. suff.
any -ve value for p with the given information, m >1. so //B//.
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Re: Math Question - Data Sufficiency [#permalink] New post 17 Sep 2008, 04:09
I don'T AGREE with puma
By (B) p could be -1 and the total outcome will be m^3=(-(-1)+ (-2)^(-1))^3=
=(1-0.5)^3<1

Could also be -2 and the total will be m^3=(2+1/4)^3>1

So b is not right- the answer is c
Re: Math Question - Data Sufficiency   [#permalink] 17 Sep 2008, 04:09
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If p is an integer and m=-p+(-2)^p, is m^3>1? 1. p is

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