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by rijul007 » Wed Jan 04, 2012 10:05 pm
karthikpandian19 wrote:If p is an integer, what is the value of p?
(1) (|p|!)^p = |p|!
(2) p^p = p^2
(1) (|p|!)^p = |p|!

this is true for p = 1 or -1

Insufficient

(2) p^p = p^2

p^p = p^2

This is true for p = 1 or 2

Combining two statements
p = 1

Option C
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by neelgandham » Thu Jan 05, 2012 1:49 am
rijul007 wrote:
karthikpandian19 wrote:If p is an integer, what is the value of p?
(1) (|p|!)^p = |p|!
(2) p^p = p^2
(1) (|p|!)^p = |p|!
this is true for p = 1 or -1
Insufficient
An addition to Rijul's solution, The value of p can also be '0' from statement 1
(|0|!)^0 = |0|!
1^0 = 1
1=1
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by santhoshsram » Thu Jan 05, 2012 7:20 pm
neelgandham wrote:
rijul007 wrote:
karthikpandian19 wrote:If p is an integer, what is the value of p?
(1) (|p|!)^p = |p|!
(2) p^p = p^2
(1) (|p|!)^p = |p|!
this is true for p = 1 or -1
Insufficient
An addition to Rijul's solution, The value of p can also be '0' from statement 1
(|0|!)^0 = |0|!
1^0 = 1
1=1
Just curious. Trying to plug 0 in (2), we get 0^0 on the LHS. Is 0^0 indeterminate or 1 on GMAT?
-- Santhosh S
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by karthikpandian19 » Thu Jan 05, 2012 7:27 pm
If it is takes ZERO then there will be 2 solutions and the answer shld be E????

Is that right?
neelgandham wrote:
rijul007 wrote:
karthikpandian19 wrote:If p is an integer, what is the value of p?
(1) (|p|!)^p = |p|!
(2) p^p = p^2
(1) (|p|!)^p = |p|!
this is true for p = 1 or -1
Insufficient
An addition to Rijul's solution, The value of p can also be '0' from statement 1
(|0|!)^0 = |0|!
1^0 = 1
1=1
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by rijul007 » Thu Jan 05, 2012 11:01 pm
karthikpandian19 wrote:If it is takes ZERO then there will be 2 solutions and the answer shld be E????

Is that right?
neelgandham wrote:
rijul007 wrote:
karthikpandian19 wrote:If p is an integer, what is the value of p?
(1) (|p|!)^p = |p|!
(2) p^p = p^2
(1) (|p|!)^p = |p|!
this is true for p = 1 or -1
Insufficient
An addition to Rijul's solution, The value of p can also be '0' from statement 1
(|0|!)^0 = |0|!
1^0 = 1
1=1
st(1) would give p= -1, 0 or 1
st(2) would give p = 1 or 2

After Combining the two statements
we get p = 1

So option C should be the ans
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by karthikpandian19 » Fri Jan 06, 2012 4:02 am
Got it

i checked OA is C
rijul007 wrote:
karthikpandian19 wrote:If it is takes ZERO then there will be 2 solutions and the answer shld be E????

Is that right?
neelgandham wrote:
rijul007 wrote:
karthikpandian19 wrote:If p is an integer, what is the value of p?
(1) (|p|!)^p = |p|!
(2) p^p = p^2
(1) (|p|!)^p = |p|!
this is true for p = 1 or -1
Insufficient
An addition to Rijul's solution, The value of p can also be '0' from statement 1
(|0|!)^0 = |0|!
1^0 = 1
1=1
st(1) would give p= -1, 0 or 1
st(2) would give p = 1 or 2

After Combining the two statements
we get p = 1

So option C should be the ans
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by [email protected] » Sun Jan 22, 2012 5:17 am
Yes the correct answer is C...

Statement 1: The possible values could be {-1, 0, 1}


Statement 2: The possible values are {1 , 2}

Using Both statements:


The correct answer comes down to C...
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by Brent@GMATPrepNow » Sun Jan 22, 2012 6:01 am
santhoshsram wrote: Just curious. Trying to plug 0 in (2), we get 0^0 on the LHS. Is 0^0 indeterminate or 1 on GMAT?
Hi santhoshsram,

That's a difficult (even contentious) question to answer.
Some will say that 0^0 = 1, and that's fine. I think it's safe to say that you don't need to worry about it. The GMAT will not test you on the value of 0^0.


For more, please see https://www.beatthegmat.com/what-is-the- ... 27026.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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