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Is M^3>=1

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Source: — Data Sufficiency |
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by srivas » Tue Sep 01, 2009 7:46 am
statement I p is even

p=2 m =-P+(-2)^P -2+(-2)^2 = -2+4 = 2
so m^3 = 8 >= 1

statement II

P^3 <= -1 suppose P^3 = -8 then P = -2
m = -P+(-2)^P
-(-2)+(-2)^-2 = 2- 1/4 = 7/4
so m^3 >= 1

it is D
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by Matmasi » Tue Sep 01, 2009 9:06 am
Unfortunately I have to disagree with B, that it seems, to me, not sufficient.
Infact, P^3 <= -1 and you suppose P^3 = -8 then P = -2.

But in this way you forget that P can be also equal to -1; let's try:
-(-1) + (-2^-1) = 1 - 1/2 = 1/2
m= 1/2
M^3 = 1/8 <1
So B is not sufficient.
Other opinions would be greatly appreciate
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by PussInBoots » Tue Sep 01, 2009 10:24 am
The asnwer is C

srivas you didn't test 0, -1/1. Those numbers usually screw everything up.
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by fruti_yum » Tue Sep 01, 2009 10:56 am
PussInBoots wrote:The asnwer is C

srivas you didn't test 0, -1/1. Those numbers usually screw everything up.
PussInBoots how do u know you have to test those specific numbers?.. I ended up picking D as well before coz i didn't check those numbers!
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by hpgmat » Thu Sep 03, 2009 11:03 pm
srivas wrote:statement I p is even

p=2 m =-P+(-2)^P -2+(-2)^2 = -2+4 = 2
so m^3 = 8 >= 1

statement II

P^3 <= -1 suppose P^3 = -8 then P = -2
m = -P+(-2)^P
-(-2)+(-2)^-2 = 2- 1/4 = 7/4
so m^3 >= 1

it is D
Sirivas, your calculation for statement 2 is wrong . Assuming P is -2, M would be 9/4. M cube still greater than 1.
just a clarification.
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Re: Is M^3>=1

by qwe12 » Fri Sep 04, 2009 1:18 am
PAB2706 wrote:P is integer, m = -P+(-2)^P, Is M^3 is greater than or equal to 1.
1) P is even.
2) P^3 is less than or equal to -1.
please post the offical answer.

(1)
consider p=0, m=1 ... m^3>=1? YES
consider p=2, m=2 ... m^3>=1? YES
consider p=4, m=12 ... m^3>=1? YES

SUFFICIENT

(2)
consider p=-1, m=1/2 ... m^3>=1? NO
consider p=-2, m=9/4 ... m^3>=1? YES

NOT SUFFICIENT

Answer (A)
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