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

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If p is an integer and m=-p+(-2)^p is m >1 1. p is even [#permalink]

07 Apr 2006, 21:48

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If p is an integer and m=-p+(-2)^p is mÂ³>1

1. p is even

2. pÂ³ ≤-1

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07 Apr 2006, 22:37

its A

in st. 2

P <= -1

putting p = -1

m = (1/2) ---> m^3 < 1

p = -2 -------> m^3 > 1

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10 Apr 2006, 11:20

It's C.

(1) is not sufficient.

p = 2 leads to m^3 = 8 > 1
p = 0 leads to m^3 = 1, which isn't > 1

(2) is not sufficient

p^3 ≤ -1 means that p itself is ≤ -1.
In the case of p = -1, m^3 = 1/8 < 1.
If p = -2, m^3 = 729/64 > 1

(1) and (2) combined are sufficient since p
is smaller than -1 and so m^3 always > 1.

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14 Apr 2006, 12:21

Thanks a lot for the tip that Zero is a even number too
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14 Apr 2006, 13:33

This is what I am familiar with now

Intergers ...............-3,-2, -1, 0, 1, 2, 3

Whole numbers 0,1,2,3,4 ................................... -ve numbers not included

Prime numbers 2,3,5 ....................................

even numbers .....-4,-2, 0, 2,4,6 .........................

Odd numbers .........-3,-1 not zero 1,3,5...................

Please correct if I am wrong or add to above and let us put out a sticky of what is considered correct by GMAT lords.

Last edited by old_dream_1976 on 14 Apr 2006, 14:02, edited 1 time in total.

[#permalink] 14 Apr 2006, 13:33

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

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

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

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