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# If p is an integer and m = -p + (12)*p is m^3 >1? (1) p

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Manager
Joined: 25 Jul 2006
Posts: 99

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If p is an integer and m = -p + (12)*p is m^3 >1? (1) p [#permalink]

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07 Aug 2006, 18:32
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If p is an integer and m = -p + (12)*p is m^3 >1?

(1) p is even
(2) p^3<1

Kudos [?]: 14 [0], given: 0

VP
Joined: 02 Jun 2006
Posts: 1257

Kudos [?]: 108 [0], given: 0

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07 Aug 2006, 18:37
Given m = -p + 12p = 11p
Is 11^3 x p^3 > 1?

S1: p is even

If p = 2, 11^3 x 8 > 1 is true.
If p =-2, 11^3 x-8 > 1 is false.

Not sufficient.

S2: p^3 < 1
As p is an integer, p < 0 i.e. p is -ve.
Therefore, 11^3xp^3 > 1 is false.
Sufficient.

Kudos [?]: 108 [0], given: 0

Intern
Joined: 22 Jan 2006
Posts: 3

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Re: DS with number properties [#permalink]

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07 Aug 2006, 22:08
2times wrote:
If p is an integer and m = -p + (12)*p is m^3 >1?

(1) p is even
(2) p^3<1

m=-p + (12)*p=11p

Is m^3>1?
statement 1: p is even
p=-2, 2,4...

not sufficient

statement 2: p^3<1
=> p must be -ve
=>m = 11p is also -ve
=>m^3<0 < 1
sufficient

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SVP
Joined: 30 Mar 2006
Posts: 1728

Kudos [?]: 102 [0], given: 0

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07 Aug 2006, 22:11
B

1) p is even

hence p can be -4, -2, 2, 4 etc
when p = -2
m = -22, m^3 <1
When p = 2
m = 26, m^3 > 1

Not suff.

2) p^3 < 1
Hence p is negative.
Then m = -ve
hence m^3 is always less than 1
Suff

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Manager
Joined: 20 Mar 2006
Posts: 200

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Re: DS with number properties [#permalink]

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11 Aug 2006, 20:45
2times wrote:
If p is an integer and m = -p + (12)*p is m^3 >1?

(1) p is even
(2) p^3<1

p={ I, 0,+,-}

(1) p=0 --> m=0--> m^3> 1 --> No
p=2--> m=22--> m^3> 1 --> Yes
BCE

(2)
p=0 --> m=0--> m^3> 1 --> No
p= -1 --> m= -11 --> m^3> 1 --> No

Hence B
Heman

Kudos [?]: 5 [0], given: 0

Re: DS with number properties   [#permalink] 11 Aug 2006, 20:45
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