Is q^b a divisor of p^a ? 1. ab 2. p is a multiple of q

Question

Is q^b a divisor of p^a ?

1. a>b

2. p is a multiple of q

All numbers are integers.

vasild · Answer

The question is if p^a divided by q^b equals an integer. (I am assuming q isn't zero)

(1) doesn't help much by itself. Obviously insufficient.

(2) tells us that p=q*x, where x is some random integer

Substitute into the stem and we now have to evaluate the following number: (q*x)^a / q^b

We can rewrite it as 

(q^a)*(x^a) / q^b = q^(a-b) * x^a

x^a is definitely an integer, and if a>=b the same is true for the other factor. This is when statement (1) somes into play, because it gives us the needed information

=> (C) is the answer.

Am I missing something?

Vasil

HIMALAYA · Answer

if a>b and p is a multiple of q, then q^b can be divisor of p^a. suppose, if a=3, b =2, and p=q=1, p^a can be divided by q^b. so I go with C.

prep_gmat · Answer

E.

a = -1 > b = -2
p=6 a multiple of q = 2.

Also, q = p can be zero.

HongHu · Answer

I would expect a real GMAT question to define q, p, a, and b positive integers. If this is the case, then C should be the correct answer.

From 2, p=mq
p^a/q^b=m^aq^a/q^b=m^aq^(a-b)
From 1, a-b>0
then p^a/q^b=m^aq^(a-b) is also an integer. In other words q^b is a divisor of p^a.

However if pqab are not defined, then we don't know if they are integers. For example, a-b might be 1/2 or something. Then the answer would be E.

AJB77 · Answer

OA is C

HongHu, I noted in the Question that all numbers are integers.

HongHu · Answer

You are very right AJB, though I would prefer they are positive integers, them being integers are the most important one and it is defined in the question. Now can 0 be a divisor to another number? :hmmm:

prep_gmat · Answer

If p,q,a,b are not positive integers, How can the answer be C?.

p=6, q=2, a=-1, b=-2 satisfy both statements and yield a 'no' answer. One can find several easy examples of 'yes' answer for a given set of p,q,a,b if all are positive and both statements are true.

HongHu · Answer

I believe that the domain is the natural numbers when we are talking about multiples and factors. However, 0 should be included. 0 is a multiple of any number, including 0 itself. In other words for this question both q and p can be zero to satisfy (2), and then we would get a "No" to the question (since 0 can't be a divisor) instead of "Yes" when they are not zeros.

Also to make sure q^b and p^a are natural numbers so that "divisor" is defined, a and b must not be negative.

Therefore for this question to be strictly defined it should say all numbers are positive integers.

Is q^b a divisor of p^a ? 1. a>b 2. p is a multiple of q

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