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M30-11

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M30-11  [#permalink]

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New post 16 Sep 2014, 00:45
2
3
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

38% (01:24) correct 62% (01:30) wrong based on 114 sessions

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Re M30-11  [#permalink]

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New post 16 Sep 2014, 00:45
3
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.


Answer: B
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Re: M30-11  [#permalink]

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New post 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.


Answer: B

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
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Re: M30-11  [#permalink]

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New post 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.


Answer: B

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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Collection of Questions:
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Re: M30-11  [#permalink]

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

or 5*3, 7*2
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Re: M30-11  [#permalink]

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New post 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
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Re: M30-11  [#permalink]

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New post 25 Dec 2017, 06:37
1
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M30-11  [#permalink]

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New post 30 Sep 2018, 13:34
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Bunuel

This question is same as the question M30-14.
Link to the question: https://gmatclub.com/forum/m30-184572.html
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Re: M30-11  [#permalink]

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New post 30 Sep 2018, 19:42
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Re: M30-11 &nbs [#permalink] 30 Sep 2018, 19:42
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