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

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 have to check whether p^3q is an integer or not.

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

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

@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?

@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

bubbliiiiiiii wrote:@manpsingh87,

I think your approach is more inclined towards finding if the given term is positive or negative integer.

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: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]

As is the case with a lot of complex DS questions, we can make our lives easier by simplifying the question.

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)!

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.

nice question...

IMO-E.