It is currently 21 Mar 2018, 06:02

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If p is an integer, then p is divisible by how many positive

Author Message
Senior Manager
Joined: 25 Nov 2006
Posts: 333
Schools: St Gallen, Cambridge, HEC Montreal
If p is an integer, then p is divisible by how many positive [#permalink]

### Show Tags

08 Feb 2009, 13:03
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If p is an integer, then p is divisible by how many positive integers?

(1) p=2^x, where x is a prime number.

(2) p=x^2, where x is a prime number.
Intern
Joined: 07 Jan 2009
Posts: 18
Location: Boston/Cleveland

### Show Tags

08 Feb 2009, 13:41
This one is kind of similar like the other one you posted I think.
1. This one should be easy to figure out, as x gets larger there will be more divisible integers so we don't know wtf the exact number is going to be
2. This one we just have to factor X^2 -> X,X -> oh wait that's it, since x is a prime number. Answer is B.

Factoring is the right terminology right?

Last edited by vladmoney on 06 Mar 2009, 13:08, edited 1 time in total.
Senior Manager
Joined: 30 Nov 2008
Posts: 482
Schools: Fuqua

### Show Tags

08 Feb 2009, 15:09
IMO B. The approach is slightly different.

The number of possible factors of a number is the product of (powers of each distinct prime factor + 1)

For ex, Possible factors of 24 can be found as follows -

$$24 = 2^3 * 3$$ ==> powers of 2 and 3 are 3, 1 ==> Possible factors of 24 are $$(3+1) * (1 + 1) = 4 * 2 = 8.$$

Now coming back to our question, it is asking what is the number of possible factors of p.

From stmt 1, it is given p = $$2^x$$. Possible factors will be x + 1. depending of x, the value keeps changing. Hence insufficient.

From Stmt 2, it is given that $$p = x^2$$ where x is a prime. So the possible factors of P will be 2 + 1 = 3. Sufficent.
SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York

### Show Tags

08 Feb 2009, 15:15
lumone wrote:
If p is an integer, then p is divisible by how many positive integers?

(1) p=2^x, where x is a prime number.

(2) p=x^2, where x is a prime number.

B.

p= x^2 = 1.x.x ( x is prime number)

P always divisible by 3 positive integers.. (1,x,x^2)
_________________

Smiling wins more friends than frowning

Re: DS- 12   [#permalink] 08 Feb 2009, 15:15
Display posts from previous: Sort by