Discussion
Suppose p is even and q is odd. Then which of the following CANNOT be an integer?
| (A) | |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 03/05/2014 09:06
I will disagree with the answer, since we choose p=2 and q=1, in the A case we will have (2+1)/1=3, which is an integer!!!!!!!!!!!!!
Posted: 03/27/2014 16:17
Natali, you are right. There was a mistake in the answer choice A. It said (p+q)/q, whereas it should have said (p+q)/p. We apologize for the error.
Posted: 04/13/2014 18:33
All of the answer choices can be an integer.
For a:
p=6, q=3. (6+3)/3=3.
For b:
p=6, q=3. (6x3)/3=6.
For c:
p=18, q=3. 18/(3^2)=2.
For a:
p=6, q=3. (6+3)/3=3.
For b:
p=6, q=3. (6x3)/3=6.
For c:
p=18, q=3. 18/(3^2)=2.
Posted: 04/21/2014 14:02
Phil, we apologize. There was a mistake in answer choice A. It should have said (p+q) / p instead of (p+q)/q.
For C, I think you made a mistake in your calculation. It is q / p^2, not p/q^2.
For C, I think you made a mistake in your calculation. It is q / p^2, not p/q^2.
Posted: 05/13/2014 07:04
In this question you state that A cannot be an answer. The equation you use in the explanation is for a different quotient: p+q/p. Answer choice A can be correct when using the correct divisor, q.
Using p=2 and q=1, then 2+1=3/1
Using p=2 and q=1, then 2+1=3/1
Posted: 05/13/2014 07:08
Reply: I found the previous comment regarding this problem. Please correct it so that others will not have the same issue.
Posted: 05/19/2014 11:45
We are doing an update for GRE app today. Thanks for the reminder, Rhonda.
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