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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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Difficulty:   75% (hard)

Question Stats: 40% (01:21) correct 60% (01:32) wrong based on 136 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 $$p$$.

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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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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.

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Intern  Joined: 10 Jul 2014
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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 V
Joined: 02 Sep 2009
Posts: 56300

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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.
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Manager  S
Joined: 17 Aug 2015
Posts: 93
GMAT 1: 650 Q49 V29 ### Show Tags

first one 7*3, 11*2

or 5*3, 7*2
Manager  G
Joined: 30 Dec 2016
Posts: 234
GMAT 1: 650 Q42 V37 GPA: 4

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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 V
Joined: 02 Sep 2009
Posts: 56300

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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.
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Retired Moderator G
Joined: 11 Aug 2016
Posts: 375

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Bunuel

This question is same as the question M30-14.
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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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Bunuel

This question is same as the question M30-14.

Thank you. Removed this one from the tests' database.
_________________ Re: M30-11   [#permalink] 30 Sep 2018, 20:42
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# M30-11

Moderators: chetan2u, Bunuel  