It is currently 22 Jan 2018, 00:30

### 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

# M30-11

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43357

### Show Tags

16 Sep 2014, 00:45
Expert's post
3
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

48% (00:59) correct 52% (00:57) wrong based on 31 sessions

### HideShow timer Statistics

If $$p$$ is a positive integer, is $$p$$ a prime number?

(1) $$p$$ and $$p+1$$ have the same number of factors.
(2) $$p-1$$ is a factor of $$p$$.
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 43357

### Show Tags

16 Sep 2014, 00:45
Expert's post
2
This post was
BOOKMARKED
Official Solution:

(1) $$p$$ and $$p+1$$ have the same number of factors.

Primes have 2 factors, 1 and itself, (the reverse is also true: if a positive integer has 2 factors, then it must be a prime). So, for the answer to the question to be YES, both $$p$$ and $$p+1$$ must be primes. Are there consecutive primes? Yes, 2 and 3.

Could we have a case when $$p$$ and $$p+1$$ have the same number of factors, and $$p$$ is NOT a prime? Yes. For example, both 14 (not a prime) and 15 have four factors. Also, both 21 (not a prime) and 22 have four factors.

Not sufficient.

(2) $$p-1$$ is a factor of $$p$$.

$$p-1$$ and $$p$$ are consecutive integers. Consecutive integers do not share any common factor but 1. Therefore, $$p-1$$ to be a factor of $$p$$, $$p-1$$ must be 1, which makes $$p$$ equal to prime number 2. Sufficient.

_________________
Intern
Joined: 10 Jul 2014
Posts: 1

### Show Tags

25 Sep 2014, 04:14
Official Solution:

(1) p and p+1 have the same number of factors.

Primes have 2 factors, 1 and itself, (the reverse is also true: if a positive integer has 2 factors, then it must be a prime). So, for the answer to the question to be YES, both p and p+1 must be primes. Are there consecutive primes? Yes, 2 and 3.

Could we have a case when p and p+1 have the same number of factors, and p is NOT a prime? Yes. For example, both 14 (not a prime) and 15 have four factors. Also, both 21 (not a prime) and 22 have four factors.

Not sufficient.

(2) p-1 is a factor of p.

p-1 and p are consecutive integers. Consecutive integers do not share any common factor but 1. Therefore, p-1 to be a factor of p, p-1 must be 1, which makes p equal to prime number 2. Sufficient.

Bunuel

BUt for condition 2.

Any prime no greater than 2 cannot have p-1 as a factor.
So dont you think that condn 2 gives both YES?NO Answer-------- hence not sufficient.
Please let me know what I m missing.
Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 43357

### Show Tags

25 Sep 2014, 06:30
Rocker69 wrote:
Official Solution:

(1) p and p+1 have the same number of factors.

Primes have 2 factors, 1 and itself, (the reverse is also true: if a positive integer has 2 factors, then it must be a prime). So, for the answer to the question to be YES, both p and p+1 must be primes. Are there consecutive primes? Yes, 2 and 3.

Could we have a case when p and p+1 have the same number of factors, and p is NOT a prime? Yes. For example, both 14 (not a prime) and 15 have four factors. Also, both 21 (not a prime) and 22 have four factors.

Not sufficient.

(2) p-1 is a factor of p.

p-1 and p are consecutive integers. Consecutive integers do not share any common factor but 1. Therefore, p-1 to be a factor of p, p-1 must be 1, which makes p equal to prime number 2. Sufficient.

Bunuel

BUt for condition 2.

Any prime no greater than 2 cannot have p-1 as a factor.
So dont you think that condn 2 gives both YES?NO Answer-------- hence not sufficient.
Please let me know what I m missing.
Thanks

Not sure I can follow you...

p-1 is a factor of p means that p-1 must be 1, thus p=2=prime.
_________________
Manager
Joined: 17 Aug 2015
Posts: 123
GMAT 1: 650 Q49 V29

### Show Tags

21 Sep 2016, 13:39
first one 7*3, 11*2

or 5*3, 7*2
Manager
Joined: 30 Dec 2016
Posts: 86

### Show Tags

25 Dec 2017, 06:33
Hi Bunuel. Merry Christmas.

I have a general question. To prove that statement 1 is insufficient is testing numbers the only way?
Thinking of numbers that can have same number of factors, that too Under the time pressure to solve a question with in 2 minutes.! ??

Regards
Math Expert
Joined: 02 Sep 2009
Posts: 43357

### Show Tags

25 Dec 2017, 06:37
1
KUDOS
Expert's post
sandysilva wrote:
Hi Bunuel. Merry Christmas.

I have a general question. To prove that statement 1 is insufficient is testing numbers the only way?
Thinking of numbers that can have same number of factors, that too Under the time pressure to solve a question with in 2 minutes.! ??

Regards

Yes, for this particular question testing values is the best way to get that (1) is not sufficient.
_________________
Re: M30-11   [#permalink] 25 Dec 2017, 06:37
Display posts from previous: Sort by

# M30-11

Moderators: chetan2u, Bunuel

 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®.