If p and q are integers and pq != (not equal to) 0, is p^3q (p power 3q) an integer?
q^3p is an integer
q^3p is greater than 0.
Substituting number for p and q is one approach. Can anyone suggest something more generalized?
[spoiler]OA: E[/spoiler]
Exponents
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- bubbliiiiiiii
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given p = q = non zero
using 1) p is positive
Using 2) q^3p > 0, p,q can be multiple set
combind:
p = 1 , q = -1 will do
p = 3 , q = -1 will not do
p = 1, q= 1 will do
E
using 1) p is positive
Using 2) q^3p > 0, p,q can be multiple set
combind:
p = 1 , q = -1 will do
p = 3 , q = -1 will not do
p = 1, q= 1 will do
E
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HSPA.
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Regards,
HSPA.
- manpsingh87
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we have to check whether p^3q is an integer or not.bubbliiiiiiii wrote:If p and q are integers and pq != (not equal to) 0, is p^3q (p power 3q) an integer?
q^3p is an integer
q^3p is greater than 0.
Substituting number for p and q is one approach. Can anyone suggest something more generalized?
[spoiler]OA: E[/spoiler]
1)q^3p is an integer, different cases under which it can become an integer.
a) q= -ve integer,p=+ve and odd integer, q^3p becomes negative integer,
b)q=-ve integer,P=+ve and even integer, q^3p becomes positive integer,
c)q=+ve integer, p=+ve (even or odd) integer q^3p positive integer
as here different cases are possible,hence 1 alone is not sufficient to answer the question.
2)q^3p is greater than zero, here following cases are possible.
a)q=-ve integer,P=+ve and even integer, q^3p becomes positive integer,
b)q=+ve integer, p=+ve (even or odd) integer q^3p positive integer
therefore 2 alone is also not sufficient to answer the question.!!
upon combining 1 and 2 we are left with two following possible cases,
a)q=-ve integer,P=+ve and even integer, q^3p becomes positive integer,
b)q=+ve integer, p=+ve (even or odd) integer q^3p positive integer
for case a p^3q is a fraction, and for case b its an integer, hence even after combining 1 and 2 we are not getting any unique solution. hence answer should be E
Last edited by manpsingh87 on Sat Apr 09, 2011 7:23 am, edited 1 time in total.
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statement 1: q^3p is an integer..Now, it can be both a positive or a negative integer..and thus, we cannot be sure of p^3q...(in case 3q is positive, p^3q will be an integer, however, if 3q<-1, p^3q won't be an integer) Thus, 1 is insufficient.
statement 2: q^3p>0..Now, if q is a positive integer, we can be sure of p^3q being an integer..however, what is q is a positive fraction..Thus, 2 also is insufficient.
Thus, E
statement 2: q^3p>0..Now, if q is a positive integer, we can be sure of p^3q being an integer..however, what is q is a positive fraction..Thus, 2 also is insufficient.
Thus, E
- bubbliiiiiiii
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@manpsingh87,
I think your approach is more inclined towards finding if the given term is positive or negative integer.
@Pankaj,
Great. This goes inline with the approach I used to solve this question.
Can you elaborate, how 1 and 2 both are insufficient?
I think your approach is more inclined towards finding if the given term is positive or negative integer.
@Pankaj,
Great. This goes inline with the approach I used to solve this question.
Can you elaborate, how 1 and 2 both are insufficient?
Regards,
Pranay
Pranay
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@bubbl(i)^8
My approach is based on the following fact: Let x be an integer;
a) (x)^(Positive integer) gives an integer
b) (x)^(negative integer) gives a fraction (unless x=1)
Employ the fact described above in both the statements given in the question. Let me know, if you need any more explanation.
Thanks
My approach is based on the following fact: Let x be an integer;
a) (x)^(Positive integer) gives an integer
b) (x)^(negative integer) gives a fraction (unless x=1)
Employ the fact described above in both the statements given in the question. Let me know, if you need any more explanation.
Thanks
- manpsingh87
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well, i elaborated all the possible scenarios, and if you could have read my post carefully you might have understood that i did it purposely to explain how depending upon q= +ve or -ve, our answer can be different...!!! any ways its your call, to go with whichever solution you want..!!!!bubbliiiiiiii wrote:@manpsingh87,
I think your approach is more inclined towards finding if the given term is positive or negative integer.
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- Stuart@KaplanGMAT
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As is the case with a lot of complex DS questions, we can make our lives easier by simplifying the question.bubbliiiiiiii wrote:If p and q are integers and pq != (not equal to) 0, is p^3q (p power 3q) an integer?
q^3p is an integer
q^3p is greater than 0.
Substituting number for p and q is one approach. Can anyone suggest something more generalized?
[spoiler]OA: E[/spoiler]
We know that p and q are both non-zero integers. So, when will p^(3q) NOT be an integer?
Well, if q is positive, then p^(3q) will always be an integer.
If q is negative, then p^(3q) will never be an integer.
So, rephrasing the question:
Is q positive?
(1) q^3p is an integer.
This tells us that p is positive, but q could still be any integer: insufficient.
(2) q^3p is greater than 0.
If p is odd, then q must be positive. However, if p is even, q could be positive or negative: insufficient.
Together: we know that p is positive, but still don't know if it's even or odd, so q could be positive or negative. Insufficient, choose (E)!
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- bubbliiiiiiii
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Thanks Stuart for such a great approach.
Actually, I have followed the approach posted be pankaj earlier.
OE suggests method of substitution.
Your method is great!! I would try and follow the approach of simplifying a question and then try to validate the options.
Actually, I have followed the approach posted be pankaj earlier.
OE suggests method of substitution.
Your method is great!! I would try and follow the approach of simplifying a question and then try to validate the options.
Regards,
Pranay
Pranay