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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: 690
Followers: 5

Kudos [?]: 30 [0], given: 0

If p is an integer and m=-p+(-2)^p is m^3>1? 1) p is [#permalink]  14 Jan 2005, 17:43
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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
Director
Joined: 19 Nov 2004
Posts: 563
Location: SF Bay Area, USA
Followers: 3

Kudos [?]: 75 [0], given: 0

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, 18:15, edited 1 time in total.
VP
Joined: 18 Nov 2004
Posts: 1441
Followers: 2

Kudos [?]: 22 [0], given: 0

"C" for me. Will explain if correct.
Director
Joined: 07 Jun 2004
Posts: 614
Location: PA
Followers: 3

Kudos [?]: 317 [0], given: 22

Nocilis 0 is an even integer and for that m^3 = 1
Director
Joined: 07 Nov 2004
Posts: 690
Followers: 5

Kudos [?]: 30 [0], given: 0

Good catch. OA is C.
Director
Joined: 19 Nov 2004
Posts: 563
Location: SF Bay Area, USA
Followers: 3

Kudos [?]: 75 [0], given: 0

Damn ! ... Yes 0 is an even number and ignored it.
It is C then
Manager
Joined: 25 Oct 2004
Posts: 249
Followers: 1

Kudos [?]: 7 [0], given: 0

yes i would go with C. Good question ......
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