Discussion
If p and q are positive integers, how many integers are larger than pq and smaller than p(q + 2)?
*This question is included in Nova Math - Problem Set A: Substitution, question #9
| (A) | 3 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 05/07/2012 23:34
I need more clarification on this. The answer doesn't make sense
Posted: 05/10/2012 02:27
You are looking for integers that are larger than pq and smaller than p(q+2); or:
pq < N < pq + 2p. How many Ns can there be?
Think this way. How many integers can there be between, say, 3 and 7? Well, they are 4, 5, and 6. There are 3 integers or 7 - 3 - 1. You have to minus 1 because you can't count 3 in it (between / larger than / smaller than are not inclusive).
By the same logic, you can say there can be pq+2p - pq -1 integers, or 2p - 1.
pq < N < pq + 2p. How many Ns can there be?
Think this way. How many integers can there be between, say, 3 and 7? Well, they are 4, 5, and 6. There are 3 integers or 7 - 3 - 1. You have to minus 1 because you can't count 3 in it (between / larger than / smaller than are not inclusive).
By the same logic, you can say there can be pq+2p - pq -1 integers, or 2p - 1.
Posted: 06/30/2012 16:24
Even using p=1and q=2, I am having issues understanding this.
1x2= 2 =pq
(2x1)-1= 1 = 2p-1
Am I missing something here or am I working it out wrong?
1x2= 2 =pq
(2x1)-1= 1 = 2p-1
Am I missing something here or am I working it out wrong?
Posted: 06/30/2012 20:38
Cef,
The question is asking how many integers between pq and (p+2)q, if you set p = 1 and q=2,
then pq = 2, and (p+2)q = 6; there are 3 integers betwen 2 and 6, and 2p-1 = 3.
The question is asking how many integers between pq and (p+2)q, if you set p = 1 and q=2,
then pq = 2, and (p+2)q = 6; there are 3 integers betwen 2 and 6, and 2p-1 = 3.
Posted: 04/03/2015 07:44
Reply: it's pq (1*2)=2' and p(q+2) or 1(2+2)=4. And the only integer between 2 &4 is 3. Is not it ?Why you change the equation to get q(p+2)=6?
Posted: 07/01/2012 21:01
Thank you, it makes sense now.
Posted: 11/27/2012 14:03
If it stipulates that p is a positive integer then why do we set p as 1, which is odd?
Posted: 11/27/2012 14:17
Ian, a positive integer can be odd or even.
Posted: 03/18/2013 12:36
What if you used p=2 and q=3? Is that wrong to do that? Cause then the answer would be three? 6
Posted: 03/19/2013 17:10
Julia, pq = 6, and pq + 2p =10. There are 3 integers between 6 and 10, or 3 = 2 (2) - 1 = 2p - 1
Posted: 08/09/2013 06:46
If you use p=1 and q=2 then pq =2 and p(q+2)=4 so the only integer between 2&4 is 3. pq+1= 1*2+1=3 that should be the answer. Not pq-1=1*2-1=1 not 3. So why is choice E not the correct answer?
Posted: 08/09/2013 06:56
O never mind I miss read the question
Posted: 08/20/2013 20:21
If p=1 and q=2, then wouldn't c also make sense? P-2= 1-2=-1, whic isn't between 2 and 4
Posted: 08/30/2013 12:58
Ashley, the question is asking for the number of integers between two boundaries, which means the number has to be a positive integer. -1 is not a positive integer.
Posted: 11/30/2013 19:41
I'm asking about choice a , why it is not the correct answer ?
Posted: 12/16/2013 17:17
Targ. Please read the explanation. It is pretty clear from the explanation that there is a scenario in which there is a number of integers different than 3, larger than pq but less than p(q+2)
Posted: 12/15/2013 10:29
Ok, it asks if p and q are positive integers. If I chose both p and q to equal 2 then pq=4 and p(q+2)=8, hence 4 < x < 8, where 2p + 1 is the only possible answer.
Posted: 12/15/2013 10:31
Scratch that last comment, I wrote it before reading the comments. Sorry
Posted: 02/10/2015 09:39
I don't understand the second part of solving this problem. Why do we need to find a solution equal to 1? Why is that significant?
Posted: 02/10/2015 09:40
Never mind, I figured it out!
Posted: 05/22/2015 12:47
I am glad you did. Did you take the test? How was it?
Posted: 10/01/2016 09:28
If we use p=4 and q=5
pq = 20
p(q+2) = 22
Hence 1 integere
HOWEVER,
D)
2p-1=9
which is not the right answer.
The answer D is not working with any positive integers.
Appreciate clarification
Thanks
pq = 20
p(q+2) = 22
Hence 1 integere
HOWEVER,
D)
2p-1=9
which is not the right answer.
The answer D is not working with any positive integers.
Appreciate clarification
Thanks
Posted: 10/01/2016 09:30
Oooops 2p-1=7
But still this makes D a wrong answer...as shown above
But still this makes D a wrong answer...as shown above
Posted: 10/01/2016 19:10
Hi Amir,
p(q + 2) = 4(5 + 2) = 4(4) = 28. Now, there are 7 integers between 20 and 28, which is what choice D generates: 2p - 1 = 2(4) - 1 = 8 - 1 = 7.
Nova Press
p(q + 2) = 4(5 + 2) = 4(4) = 28. Now, there are 7 integers between 20 and 28, which is what choice D generates: 2p - 1 = 2(4) - 1 = 8 - 1 = 7.
Nova Press