 |
| Author |
Message |
|
andrew.k.john
|
Post subject: If x^3 - x = p, and x is odd.....  Posted: Sat Jun 13, 2009 1:16 pm |
|
 |
| Course Students |
|
|
Posts: 13
|
|
Question regarding an example from GMAT Strat. Guide: Number Properties (12th ed.), page 151.
If x^3 - x = p, and x is odd, is p divisible by 24?
The explanation is quite thorough in offering a solution for all x>1, but does not
consider x <= 1. Here are two counterexamples...
If x=1, the p=0. Can 0 be divisible by 24? If in GMAT land, 0 is divisible by all numbers, then I have not yet covered that reading.
If x= (-1), then p= (-2).
I believe the original statement requires a statement that x>1 for the book's explanation to hold merit.
Thoughts?.....
|
|
  |
|
|
|
deshpande.harsha
|
Post subject: Re: If x^3 - x = p, and x is odd..... Posted: Thu Jun 25, 2009 2:37 pm |
|
|
| Students |
|
|
Posts: 2
|
|
Hi Andrew,
If this is a DATA Suff. question then the answer has to be (E)..neither of the statements can answer the question, since they have not mentioned that x is posotive odd...So we can assume negative odd numbers for x anf fail the test.
|
|
| |
|
|
|
muktarashmi
|
Post subject: Re: If x^3 - x = p, and x is odd..... Posted: Thu Jul 02, 2009 2:41 pm |
|
|
| Course Students |
|
|
Posts: 15
|
|
Hi Andrew it works like this
X(X^2 - 1)= X.(X-1).(x+1)
Since X is odd
It has to be
3.2.4..so, P is not Divisible
23.24.22= P is Divisible
So, answer will be not sufficient
|
|
| |
|
|
|
andrew.k.john
|
Post subject: Re: If x^3 - x = p, and x is odd..... Posted: Fri Jul 03, 2009 6:06 pm |
|
|
| Course Students |
|
|
Posts: 13
|
|
I've read my original post several times, and am scratching my head as to why I've received two responses assuming this to be a data sufficiency problem. This is not a DS problem. It is not a PS problem either. It is an example from the "number properties" book, as mentioned in my original post. In any case, thank you for the two responses.
Deshpande: I believe you assumed my two counterexamples to be the Facts of a DS problem. If so, thank you for the answer.
Muktarashmi: Your explanation is close, but fails in one important place. 2*3*4 is not only divisible of 24, it IS 24, thus negating your claim of Insufficiency.
If either of you have the 12th Editions of the Number Properties strategy guides, I would be interested in hearing your thoughts on the example covered on page 151.
|
|
| |
|
|
|
andrew.k.john
|
Post subject: Re: If x^3 - x = p, and x is odd..... Posted: Fri Jul 03, 2009 6:21 pm |
|
|
| Course Students |
|
|
Posts: 13
|
|
Additionaly, I just noticed an error in my original post.
If x= -1, the p=0, not -2. Bad math by me, sorry.
I believe my question is still valid, as I'm curious if we can say 0 is divisible by 24 (or any number).
|
|
| |
|
|
|
Ben Ku
|
Post subject: Re: If x^3 - x = p, and x is odd..... Posted: Fri Jul 31, 2009 1:11 am |
|
|
| ManhattanGMAT Staff |
|
|
Posts: 824
|
|
Hi Andrew,
Thanks for the question. This is from the 4th Edition of our Number Properties Strategy Guide, page 151. I'll rewrite the question:
If x^3 - x = p, and x is odd, is p divisible by 24?
Because x^3 - x can be factored into (x - 1)(x)(x +1), or three consecutive integers, then we know:
(1) (x - 1) and (x + 1) are both even, and thus are both multiples of 2
(2) (x - 1) and (x + 1) are consecutive even integers, so one of them must be a multiple of 4
(3) one of (x - 1), x, or (x + 1) is a multiple of 3
Therefore, p is divisible by 2, 3, and 4, so yes, p is divisible by 24.
If x = 1, then p = 0. As you pointed out, 0 is divisible by 24, so it is not a counterexample.
If x = -1, then p = 0. [(-1)^3 - (-1) = (-1 + 1) = 0] This is also not a counterexample.
Therefore it's unnecessary to say x > 1 in order for the explanation to be consistent with the answer.
I hope that helps!
_________________
Ben Ku
Instructor
ManhattanGMAT
|
|
| |
|
|
|
ankitp
|
Post subject: Re: If x^3 - x = p, and x is odd..... Posted: Tue Dec 01, 2009 3:22 am |
|
|
| Students |
|
|
Posts: 23
|
Ben Ku wrote: Hi Andrew,
Thanks for the question. This is from the 4th Edition of our Number Properties Strategy Guide, page 151. I'll rewrite the question:
If x^3 - x = p, and x is odd, is p divisible by 24?
Because x^3 - x can be factored into (x - 1)(x)(x +1), or three consecutive integers, then we know:
(1) (x - 1) and (x + 1) are both even, and thus are both multiples of 2
(2) (x - 1) and (x + 1) are consecutive even integers, so one of them must be a multiple of 4
(3) one of (x - 1), x, or (x + 1) is a multiple of 3
Therefore, p is divisible by 2, 3, and 4, so yes, p is divisible by 24.
If x = 1, then p = 0. As you pointed out, 0 is divisible by 24, so it is not a counterexample.
If x = -1, then p = 0. [(-1)^3 - (-1) = (-1 + 1) = 0] This is also not a counterexample.
Therefore it's unnecessary to say x > 1 in order for the explanation to be consistent with the answer.
I hope that helps! The line : (2) (x - 1) and (x + 1) are consecutive even integers, so one of them must be a multiple of 4 Where is this rule coming from? I see it mentioned in the guides solution, but I don't see any more detail. What is the general expanded rule ?
|
|
| |
|
|
|
StaceyKoprince
|
Post subject: Re: If x^3 - x = p, and x is odd..... Posted: Mon Feb 22, 2010 5:01 pm |
|
|
| ManhattanGMAT Staff |
|
|
Posts: 5788
Location: San Francisco
|
|
Great question. So, if you know x is an integer, then (x-1), x, and (x+1) are consecutive integers, right?
Okay, IF we know that x is odd, then we also know that (x-1) is even and so is (x+1). With me so far? (Test with real numbers if you're not sure.)
Now, any number that is even is (by definition) divisible by 2. So, a product that includes that number would also be divisible by 2.
In this case, we have two even numbers, and each even number is divisible by 2. So we actually have two 2s as factors of the product (x-1)(x)(x+1). two 2s = 4, so the product is therefore divisible by 4. Again, test it out with some real numbers to see how this works.
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
|
|
| |
|
|
|
sitsmeagain
|
Post subject: Re: If x^3 - x = p, and x is odd..... Posted: Thu Jun 30, 2011 5:30 am |
|
|
| Students |
|
|
Posts: 2
|
Ben Ku wrote:
If x = 1, then p = 0. As you pointed out, 0 is divisible by 24, so it is not a counterexample.
If x = -1, then p = 0. [(-1)^3 - (-1) = (-1 + 1) = 0] This is also not a counterexample.
Therefore it's unnecessary to say x > 1 in order for the explanation to be consistent with the answer.
Hello, I see that this post if old, but i am new to the new forum and brushing up my skills so that i can do well in GMAT. I have a query regarding the explanation. Inline with the quoted text, can i say that 0 is divisible by any number? for ex: If x = 1, then p = 0. So,0 is divisible by 24. In other words, if i am testing number so that the value after simplification becomes zero. I can say that 0 is divisible by that number.. Regards, Mustu
|
|
| |
|
|
|
jnelson0612
|
Post subject: Re: If x^3 - x = p, and x is odd..... Posted: Sat Jul 02, 2011 10:39 pm |
|
|
| ManhattanGMAT Staff |
|
|
Posts: 1618
|
|
Yes, zero is divisible by any number.
_________________
Jamie Nelson
ManhattanGMAT Instructor
|
|
| |
|
|
|
Users browsing this forum: No registered users and 0 guests |
| |
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum
|
|
|
|
