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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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Director
Joined: 07 Nov 2004
Posts: 683

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If p is an integer and m=-p+(-2)^p is m^3>1? 1) p is [#permalink]

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14 Jan 2005, 18:43
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If p is an integer and m=-p+(-2)^p is m^3>1?

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

Kudos [?]: 162 [0], given: 0

Director
Joined: 19 Nov 2004
Posts: 556

Kudos [?]: 272 [0], given: 0

Location: SF Bay Area, USA

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14 Jan 2005, 19:11
1) p is even.

m=-p+(-2)^p

If P were negative even integer, -p will be positive and (-2)^P will be positive and the minimum value of m will be when p is -2, m = -(-2) + (-2)^-2, which = 1.75

If P were positive even integer, the minimum value of m will be when p is 2, m = -2 + (-2)^2, which is = 2

so m^3 > 1 always
Ok

2)p^3 <= -1
This means that p is a negative integer
For for p=-1, m =0.5, but p = -2, m= 2.25
so m^3 is not always >1
Nope

A)

Last edited by nocilis on 14 Jan 2005, 19:15, edited 1 time in total.

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VP
Joined: 18 Nov 2004
Posts: 1431

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14 Jan 2005, 19:13
"C" for me. Will explain if correct.

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Director
Joined: 07 Jun 2004
Posts: 610

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Location: PA

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14 Jan 2005, 19:14
Nocilis 0 is an even integer and for that m^3 = 1

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Director
Joined: 07 Nov 2004
Posts: 683

Kudos [?]: 162 [0], given: 0

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14 Jan 2005, 19:15
Good catch. OA is C.

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Director
Joined: 19 Nov 2004
Posts: 556

Kudos [?]: 272 [0], given: 0

Location: SF Bay Area, USA

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14 Jan 2005, 19:20
Damn ! ... Yes 0 is an even number and ignored it.
It is C then

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Manager
Joined: 25 Oct 2004
Posts: 246

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15 Jan 2005, 14:57
yes i would go with C. Good question ......

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15 Jan 2005, 14:57
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