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# if p is an integer and m=-p+(-2)^p is m^3>1 1. p is even

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

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08 Nov 2006, 03:17
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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
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08 Nov 2006, 03:48
E for me
If yes will explain later
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08 Nov 2006, 03:50
I got (C)

m=-p+(-2)^p

m^3 > 1
<=> m > 1 ?

From 1
o if p=0, then m = 1
o if p=2, then m = -2 + (-2)^(2)= 2 > 1

INSUFF

From 2
p^3 <= -1
<=> p <= -1

o if p=-1, then m = 1 - 2 = -1 (< 1)
o if p=-2, then m = 2 + 1/4 (> 1)

Combining 1 & 2
Even can be negative. That means p = -2*k and k > 0.

Thus,
m=-p+(-2)^p
<=> m = 2*k + (4)^(-k) > 1 as k > 0

SUFF.
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08 Nov 2006, 04:23
Fig wrote:
I got (C)

m=-p+(-2)^p

m^3 > 1
<=> m > 1 ?

From 1
o if p=0, then m = 1
o if p=2, then m = -2 + (-2)^(2)= 2 > 1

INSUFF

From 2
p^3 <= -1
<=> p <= -1

o if p=-1, then m = 1 - 2 = -1 (< 1)
o if p=-2, then m = 2 + 1/4 (> 1)

Combining 1 & 2
Even can be negative. That means p = -2*k and k > 0.

Thus,
m=-p+(-2)^p
<=> m = 2*k + (4)^(-k) > 1 as k > 0

SUFF.

Statement 1:

Smallest even number is 2.

If we substitute 2 for p, the value of m is 2.

2 is the least value of m. m^3 >1

Statement 1 is sufficient

Statement 2:

p^3 is less than or equal to -1 means, p is less than or equal to -1.

If we take p as -1, then m^3 is less than 1
If we take p as -3, then m^3 is greater than 1

So, statement 2 is not sufficient.

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08 Nov 2006, 04:28
ak_idc wrote:
Fig wrote:
I got (C)

m=-p+(-2)^p

m^3 > 1
<=> m > 1 ?

From 1
o if p=0, then m = 1
o if p=2, then m = -2 + (-2)^(2)= 2 > 1

INSUFF

From 2
p^3 <= -1
<=> p <= -1

o if p=-1, then m = 1 - 2 = -1 (< 1)
o if p=-2, then m = 2 + 1/4 (> 1)

Combining 1 & 2
Even can be negative. That means p = -2*k and k > 0.

Thus,
m=-p+(-2)^p
<=> m = 2*k + (4)^(-k) > 1 as k > 0

SUFF.

Statement 1:

Smallest even number is 2.

If we substitute 2 for p, the value of m is 2.

2 is the least value of m. m^3 >1

Statement 1 is sufficient

Statement 2:

p^3 is less than or equal to -1 means, p is less than or equal to -1.

If we take p as -1, then m^3 is less than 1
If we take p as -3, then m^3 is greater than 1

So, statement 2 is not sufficient.

0 is an even as well .... It's inside the OG too

Defintion :
even number is an integer (positive whole numbers, zero, and negative whole numbers) divisible by 2 (with no remainder).

http://en.wikipedia.org/wiki/Even_and_odd_numbers
VP
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08 Nov 2006, 04:40
Fig wrote:
ak_idc wrote:
Fig wrote:
I got (C)

m=-p+(-2)^p

m^3 > 1
<=> m > 1 ?

From 1
o if p=0, then m = 1
o if p=2, then m = -2 + (-2)^(2)= 2 > 1

INSUFF

From 2
p^3 <= -1
<=> p <= -1

o if p=-1, then m = 1 - 2 = -1 (< 1)
o if p=-2, then m = 2 + 1/4 (> 1)

Combining 1 & 2
Even can be negative. That means p = -2*k and k > 0.

Thus,
m=-p+(-2)^p
<=> m = 2*k + (4)^(-k) > 1 as k > 0

SUFF.

Statement 1:

Smallest even number is 2.

If we substitute 2 for p, the value of m is 2.

2 is the least value of m. m^3 >1

Statement 1 is sufficient

Statement 2:

p^3 is less than or equal to -1 means, p is less than or equal to -1.

If we take p as -1, then m^3 is less than 1
If we take p as -3, then m^3 is greater than 1

So, statement 2 is not sufficient.

0 is an even as well .... It's inside the OG too

Defintion :
even number is an integer (positive whole numbers, zero, and negative whole numbers) divisible by 2 (with no remainder).

http://en.wikipedia.org/wiki/Even_and_odd_numbers

Oh...Thank you so much Fig.

I better brush up my numbers basics.
_________________

The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.

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08 Nov 2006, 04:46
ak_idc wrote:

Oh...Thank you so much Fig.

I better brush up my numbers basics.

U are welcome

Sometimes is so easy to forget ... and I'm not the last one for it ...
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08 Nov 2006, 08:42
Fig wrote:
I got (C)

m=-p+(-2)^p

m^3 > 1
<=> m > 1 ?

From 1
o if p=0, then m = 1
o if p=2, then m = -2 + (-2)^(2)= 2 > 1

INSUFF

From 2
p^3 <= -1
<=> p <= -1

o if p=-1, then m = 1 - 2 = -1 (< 1)
o if p=-2, then m = 2 + 1/4 (> 1)

Combining 1 & 2
Even can be negative. That means p = -2*k and k > 0.

Thus,
m=-p+(-2)^p
<=> m = 2*k + (4)^(-k) > 1 as k > 0

SUFF.

I think it's B.

From 2
p^3 <= -1
<=> p <= -1

o if p=-1, then m = 1 - (2^-1) = 1 - 0.5 (> 1)
SUFF
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08 Nov 2006, 14:42
hosam wrote:

I think it's B.

From 2
p^3 <= -1
<=> p <= -1

o if p=-1, then m = 1 - (2^-1) = 1 - 0.5 (> 1)
SUFF

My calculus was not good (thanks to correct it ) but also yours ... Indeed, the conclusion of the case is correct

o if p=-1, then m = 1 - (2^-1) = 1 - 0.5 = 0.5 (< 1)

Without knowning whether p is even, we alternate the sign of m. That means we are < 1 when p is odd.
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08 Nov 2006, 23:34
oa is C.

good reminder that zero is an even integer.
tx.
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