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M30-14

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M30-14 [#permalink]

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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

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Official Solution:


If \(p\) is a positive integer, is \(p\) a prime number?

(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, for \(p-1\) 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: M30-14 [#permalink]

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Bunuel, can you please advise with 2 mins/question how do we quickly determine that 14 and 15 or 21 and 22 could satisfy condition A? Although I did it correctly, I took more than 4 mins to solve this question.
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Re: M30-14 [#permalink]

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New post 04 Jun 2015, 23:06
p2bhokie wrote:
Bunuel, can you please advise with 2 mins/question how do we quickly determine that 14 and 15 or 21 and 22 could satisfy condition A? Although I did it correctly, I took more than 4 mins to solve this question.

yes this is a valid question..please explain sir.
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harshalnamdeo88 wrote:
p2bhokie wrote:
Bunuel, can you please advise with 2 mins/question how do we quickly determine that 14 and 15 or 21 and 22 could satisfy condition A? Although I did it correctly, I took more than 4 mins to solve this question.

yes this is a valid question..please explain sir.


You should spend some time and TEST values.
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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

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Re: M30-14 [#permalink]

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New post 10 Jan 2016, 12:12
Bunuel wrote:
Official Solution:


If \(p\) is a positive integer, is \(p\) a prime number?

(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, for \(p-1\) to be a factor of \(p\), \(p-1\) must be 1, which makes \(p\) equal to prime number 2. Sufficient.


Answer: B


What if P=1 than P-1=0 won't this make statement 2 insufficient?
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Re: M30-14 [#permalink]

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New post 10 Jan 2016, 12:14
rhio wrote:
Bunuel wrote:
Official Solution:


If \(p\) is a positive integer, is \(p\) a prime number?

(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, for \(p-1\) to be a factor of \(p\), \(p-1\) must be 1, which makes \(p\) equal to prime number 2. Sufficient.


Answer: B


What if P=1 than P-1=0 won't this make statement 2 insufficient?


0 is not a factor of any integer.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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.


What are GMAT Club Tests?
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M30-14 [#permalink]

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New post 17 Apr 2017, 05:34
Bunuel wrote:
harshalnamdeo88 wrote:
p2bhokie wrote:
Bunuel, can you please advise with 2 mins/question how do we quickly determine that 14 and 15 or 21 and 22 could satisfy condition A? Although I did it correctly, I took more than 4 mins to solve this question.

yes this is a valid question..please explain sir.


You should spend some time and TEST values.


You can at least think about the rule that the number of factors is the multiplication of the possibilities of the powers of its prime factors.

e.g. \(14=2^1*7^1\), power possibilities are \(2^0, 2^1\) and \(7^0, 7^1\), which multiplicates in \(2*2=4 factors\)

e.g. \(15=3^1*5^1\), power possibilities are \(3^0, 3^1\) and \(5^0, 5^1\), which multiplicates in \(2*2=4 factors\)
M30-14   [#permalink] 17 Apr 2017, 05:34
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