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If p is an integer and m= -p + (-2)^p, is m^3 >= -1

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

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If p is an integer and \(m= -p + (-2)^p\), is \(m^3 \geq -1\)

(1) p is even

(2) \(p^3 \leq -1\)
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Jul 2016, 10:38, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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

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New post 06 Jul 2010, 23:19
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Hi,

For condition 1, lets plug answers,
p=2, m= -2 + (-2)^2 = 2
p=-2, m= -(-2) + (-2)^-2 = 2 + 0.25
p=0, m= 0 +1
So condition 1 is sufficient to Answer the Q

For condition 2,
P3<=-1, hence p<=-1
Lets again plug answers,
p=-2, m= 2 + 0.25
p=-3, m= 3 - 0.125
p=-1, m= 1 - 0.5
So again condition 2 is sufficient to answer the question.

Hope the above was useful.

regards,
Jack

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

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New post 08 Jul 2010, 09:52
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St-1 : P is even
M= -p+ (-2)^p
If P is positive, (-2)^p will be positive and greater than P,
So M will be positive and m^3 >= -1

If P is negative, -p will be positive and (-2)^p will be 1/(-2)^-p which is less than 1
So M will be positive and m^3 >=1, SUFFICIENT

St-2 : P^3 <= -1
Which implies P <= -1
So –P is positive.
(-2)^p will be less than 1

So M will be positive and m^3 >= 1, SUFFICIENT

SO Ans-D
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Re: If p is an integer and m= -p + (-2)^p, is m^3 >= -1 [#permalink]

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New post 08 Jul 2010, 10:42
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Bull78 wrote:
I can´t really see the answer!

If p is an integer and \(m= -p + (-2)^p\), is \(m^3>= -1\)

1) p is even

2) \(p^3<= -1\)

Many thanks to all!!


Is \(m^3>= -1\) mean that is \(m= -1,0,1,2,....\). So if we can find the value of \(m\) we can answer the question.

Stmt \(1\): \(p\) is even. If \(p\) is even then there are two cases, whether \(p\) is positive or negative. If p is positive, then \(-p\) is negative. At the same time \(-2^p\) will be positive & adding \(-p\) & \(-2^p\) will always be positive. ( You can plug in the numbers & check). Second case, if \(p\) is negative, then \(-p\) will be positive & \(-2^p\) will result in a fraction. In this case again \(m\) will be greater than \(-1\) & we can say that stmt 1 is sufficient.

Stmt \(2\): \(p^3<= -1\). It shows that \(p<0\), means \(p\) is negative. If \(p\) is negative then \(-p\) will be positive & \(-2^p\) will result in a fraction. The addition of \(-p\) & \(-2^p\) will give the positive result. Hence Stmt 2 is also sufficient.

Answer "D". Kindly correct if I am wrong.
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Re: If p is an integer and m= -p + (-2)^p, is m^3 >= -1 [#permalink]

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New post 09 Jul 2010, 20:17
Bull78 wrote:
I can´t really see the answer!

If p is an integer and \(m= -p + (-2)^p\), is \(m^3>= -1\)

1) p is even

2) \(p^3<= -1\)

Many thanks to all!!










As always my friends, you made a great job... Correct answer D

Thanks pals..

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

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Re: If p is an integer and m= -p + (-2)^p, is m^3 >= -1   [#permalink] 20 Sep 2017, 05:16
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