GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Sep 2018, 04:31

### 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

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49252
If p is an integer, then p is divisible by how many positive integers?  [#permalink]

### Show Tags

22 Aug 2018, 01:19
1
4
00:00

Difficulty:

25% (medium)

Question Stats:

84% (01:03) correct 16% (01:36) wrong based on 49 sessions

### HideShow timer Statistics

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.

_________________
Manager
Joined: 18 Jul 2018
Posts: 190
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
If p is an integer, then p is divisible by how many positive integers?  [#permalink]

### Show Tags

Updated on: 22 Aug 2018, 01:27
1
From statement 1: P = 2^x. X can be 2,3,5,7 etc. Then P will be 4,8,32,128. Therefore the number of positive integers is not same. Hence 1 is insufficient.

From statement 2: P = x^2. X is a prime. Then X can take values 2,3,5,7 etc..
Then P = 4,9,25,49 and so on. In all such cases P has only 3 positive integers.
Example 4 has 1,2 and 4 as factors.
9 has 1,3 and 9 has factors.
Hence B is sufficient.

Posted from my mobile device

Originally posted by Afc0892 on 22 Aug 2018, 01:26.
Last edited by Afc0892 on 22 Aug 2018, 01:27, edited 1 time in total.
Director
Joined: 31 Oct 2013
Posts: 571
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
If p is an integer, then p is divisible by how many positive integers?  [#permalink]

### Show Tags

22 Aug 2018, 01:26
Bunuel 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.

we are looking for the number of factors of p.

Statement 1: $$p = 2^x$$ , where x is prime. check it.

$$p = 2^2 or 2^3 or 2^5$$

when $$2^2$$, number of factors will be 2 + 1 = 3

when $$2^3$$ , number of factors will be 3 + 1 = 4.

So, Not sufficient.

Statement 2:$$p = x^2$$, x is a prime.

$$2^2 or 3^2 or 5^2$$.

when $$2^2$$ , factors = 2 + 1 = 3

when $$3^2$$ factors = 2 + 1 = 3.

Sufficient.

Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 848
WE: Supply Chain Management (Energy and Utilities)
Re: If p is an integer, then p is divisible by how many positive integers?  [#permalink]

### Show Tags

22 Aug 2018, 01:34
Bunuel 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.

St1:- p = 2^x, where x is a prime number
Possible values of p: $$2^2,2^3,2^5,2^7$$
No of factors of p: 3,4,6,8 etc.
Insufficient

St2:- p = x^2, where x is a prime number
Here x can't be factorized further and power of x is a constant.
Hence, the no of factors is constant i.e, 3.
Sufficient.

Ans. (B)
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

Veritas Prep and Orion Instructor
Joined: 26 Jul 2010
Posts: 264
If p is an integer, then p is divisible by how many positive integers?  [#permalink]

### Show Tags

22 Aug 2018, 10:53
1
Top Contributor
For those who know the "Unique Factors Trick" statement 2 should be really quick without any number picking. In the Unique Factors Trick you break a number down to its primes and express them as exponents (so like 12 = 2^2 * 3^1), then add 1 to each exponent and multiply the exponents together (so with 12 = 2^2 * 3^1 you'd add 1 to each exponent and have (2+1)(1+1) = 3 * 2 = 6 factors (and those factors are 1, 2, 3, 4, 6, and 12).

Well here even if you don't know x, you do know that it's prime, so you've already taken the first step of the Unique Factors Trick: you have its prime base, x, taken to an exponent. Just add one to that exponent and you're set. x^2 --> 2 + 1 factors = 3 factors.

For this one, too, it's not a bad idea to know the "rule" of statement 1 in reverse: if an integer has exactly three unique factors, it's the square of a prime number.
_________________

Brian

Save \$100 on live Veritas Prep GMAT Courses and Admissions Consulting

Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Veritas Prep Reviews

If p is an integer, then p is divisible by how many positive integers? &nbs [#permalink] 22 Aug 2018, 10:53
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.