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M30-14

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M30-14  [#permalink]

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New post 16 Sep 2014, 01:45
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

53% (01:11) correct 47% (01:30) wrong based on 176 sessions

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Re M30-14  [#permalink]

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New post 16 Sep 2014, 01:45
2
1
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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New post 30 Sep 2014, 16:26
1
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 05 Jun 2015, 00:06
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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Re: M30-14  [#permalink]

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New post 05 Jun 2015, 04:54
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Re: M30-14  [#permalink]

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New post 10 Jan 2016, 13:12
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, 13:14
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.
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M30-14  [#permalink]

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New post 17 Apr 2017, 06:34
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 \(2*2=4 factors\)
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Re: M30-14  [#permalink]

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New post 05 Mar 2018, 03:42
Hi Bunuel,

Can you recap the rules for 0 again please?

I also got this question wrong because I thought of p=1 and p-1 = 0 for statement 2.

So in effect 0 is a multiple of all numbers but not a factor of any number?

Best wishes,

Tosin
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Re: M30-14  [#permalink]

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New post 05 Mar 2018, 04:28
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Re: M30-14  [#permalink]

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New post 06 Mar 2018, 05:20
If \(p\) is a positive integer, is \(p\) a prime number?


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

Let p =2 & p+1 =3.............P is prime..............Answer is Yes

Let P = 21 & p+1=22..........P is Not Prime........Answer is NO

( For clarification: factors of 21: 1,3,7,21 & factors of 22: 1,2,11,22)..the each have 4 factors)

Insufficient

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

This means that the number before p is a factor of P. This happens only in one case when P = 2.

Then P is prime = 2

Sufficient

Answer: B

Note: Two consecutive numbers have no common factor except 1.
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Re: M30-14   [#permalink] 06 Mar 2018, 05:20
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