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M31-32

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How many different prime factors does positive integer \(n\) have?

(1) \(44 < n^2 < 99\). This implies that \(n\) can be 7, 8, or 9. Each of these numbers has 1 prime: 7, 2, and 3, respectively. Sufficient.

(2) \(8n^2\) has twelve factors. For \(8n^2=2^3n^2\) to have twelve factors \(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors \(= (3+1)(2+1)=12\). Sufficient.


Answer: D
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New post 09 Nov 2015, 15:40
Bunuel wrote:
Official Solution:


How many different prime factors does positive integer \(n\) have?

(1) \(44 < n^2 < 99\). This implies that \(n\) can be 7, 8, or 9. Each of these numbers have 1 prime: 7, 2, and 3, respectively. Sufficient.

(2) \(8n^2\) has twelve factors. For \(8n^2=2^3n^2\) to have twelve factors \(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors \(= (3+1)(2+1)=12\). Sufficient.


Answer: D


Hi,

I don't understand the explanation for why (2) is sufficient. Why must "n" be a prime, and how do you get "(3+1)(2+1)=12"?

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New post 09 Nov 2015, 23:33
anders112 wrote:
Bunuel wrote:
Official Solution:


How many different prime factors does positive integer \(n\) have?

(1) \(44 < n^2 < 99\). This implies that \(n\) can be 7, 8, or 9. Each of these numbers have 1 prime: 7, 2, and 3, respectively. Sufficient.

(2) \(8n^2\) has twelve factors. For \(8n^2=2^3n^2\) to have twelve factors \(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors \(= (3+1)(2+1)=12\). Sufficient.


Answer: D


Hi,

I don't understand the explanation for why (2) is sufficient. Why must "n" be a prime, and how do you get "(3+1)(2+1)=12"?


Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
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New post 21 Jun 2016, 19:20
Why must N be a prime number? I understand that N greater than anything but 3 will result in more than 12 factors. But, just wondering if there is a trick in the number of factors which tells you that N must be a prime number?

Thanks in advance!

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ruggerkaz wrote:
Why must N be a prime number? I understand that N greater than anything but 3 will result in more than 12 factors. But, just wondering if there is a trick in the number of factors which tells you that N must be a prime number?

Thanks in advance!


If n is not a prime then 2^3*n^2 will have more than 12 factors. For example, if n is say 15, then the number of factors of 2^3*n^2 = 2^3*(3*5) will be (3+1)(1+1)(1+1)=16. Only if n is prime 2^3*n^2 will have 12 factors.

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New post 23 Jun 2016, 03:21
Hii bunuel sir,
in statement 2 n can be 2^8 and we can still get 12 factors .in that case we have only one different prime factor.is my thinking correct or wrong?

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Hii bunuel sir,
in statement 2 n can be 2^8 and we can still get 12 factors .in that case we have only one different prime factor.is my thinking correct or wrong?


I think you mean that n can be 2^4. Still in this case n would have only one prime.
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New post 23 Jun 2016, 04:07
Is it fair to say the square of all prime numbers have 3 factors?

17- It has 2 factors: 1 & 17
(17)^2- 3 factors since (2+1)=3

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New post 23 Jun 2016, 04:07
Is it fair to say the square of all prime numbers have 3 factors?

17- It has 2 factors: 1 & 17
(17)^2- 3 factors since (2+1)=3

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New post 23 Jun 2016, 07:01
ruggerkaz wrote:
Is it fair to say the square of all prime numbers have 3 factors?

17- It has 2 factors: 1 & 17
(17)^2- 3 factors since (2+1)=3


Yes, the square of a prime has 3 factors: (prime)^2 has (2+1)=3 factors.

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M31-32 [#permalink]

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New post 23 Jun 2016, 23:40
Bunuel wrote:
Official Solution:


How many different prime factors does positive integer \(n\) have?

(1) \(44 < n^2 < 99\). This implies that \(n\) can be 7, 8, or 9. Each of these numbers has 1 prime: 7, 2, and 3, respectively. Sufficient.

(2) \(8n^2\) has twelve factors. For \(8n^2=2^3n^2\) to have twelve factors \(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors \(= (3+1)(2+1)=12\). Sufficient.


Answer: D



Hi Bunuel,


It is really high quality one but I have a question.

Statement 2 proves that n has one prime number. However, if n=2 then 8n^2= 2^5 which does not yield 12 factors. Any other prime number stratifies Statement 2 but I think statement 2 contradicts statement 1 which has 2 among possible answers. My point is that the two-statements should yield same answer.

What do you think???

Thanks

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New post 27 Jun 2016, 08:34
Mo2men wrote:
Bunuel wrote:
Official Solution:


How many different prime factors does positive integer \(n\) have?

(1) \(44 < n^2 < 99\). This implies that \(n\) can be 7, 8, or 9. Each of these numbers has 1 prime: 7, 2, and 3, respectively. Sufficient.

(2) \(8n^2\) has twelve factors. For \(8n^2=2^3n^2\) to have twelve factors \(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors \(= (3+1)(2+1)=12\). Sufficient.


Answer: D



Hi Bunuel,


It is really high quality one but I have a question.

Statement 2 proves that n has one prime number. However, if n=2 then 8n^2= 2^5 which does not yield 12 factors. Any other prime number stratifies Statement 2 but I think statement 2 contradicts statement 1 which has 2 among possible answers. My point is that the two-statements should yield same answer.

What do you think???

Thanks


There is a common values for (1) and (2), so the statements do not contradict.
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New post 09 Jul 2016, 03:15
I think this is a high-quality question and I agree with explanation.

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New post 11 Oct 2016, 13:27
statement 2 seems insufficient to me
there is also a case where n = 2^4
thus 8*(n^2) = 2^11
thus yielding 12 factors but only one prime factor as 2.

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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please explain, how we got the conclusion that 'n' will be a prime number if the number of factors is 12.

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New post 17 Jul 2017, 09:37
Chets25 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please explain, how we got the conclusion that 'n' will be a prime number if the number of factors is 12.


If n is not a prime then 2^3*n^2 will have more than 12 factors. For example, if n is say 15, then the number of factors of 2^3*n^2 = 2^3*(3*5) will be (3+1)(1+1)(1+1)=16. Only if n is prime 2^3*n^2 will have 12 factors.
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New post 17 Jul 2017, 17:18
Bunuel wrote:
Chets25 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please explain, how we got the conclusion that 'n' will be a prime number if the number of factors is 12.


If n is not a prime then 2^3*n^2 will have more than 12 factors. For example, if n is say 15, then the number of factors of 2^3*n^2 = 2^3*(3*5) will be (3+1)(1+1)(1+1)=16. Only if n is prime 2^3*n^2 will have 12 factors.


Thanks Bunuel :)

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Re: M31-32 [#permalink]

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New post 26 Jul 2017, 08:49
Bunuel wrote:
Official Solution:


How many different prime factors does positive integer \(n\) have?

(1) \(44 < n^2 < 99\). This implies that \(n\) can be 7, 8, or 9. Each of these numbers has 1 prime: 7, 2, and 3, respectively. Sufficient.

(2) \(8n^2\) has twelve factors. For \(8n^2=2^3n^2\) to have twelve factors \(n\) must be a prime: \(2^3*(prime)^2\) --> number of factors \(= (3+1)(2+1)=12\). Sufficient.


Answer: D


hi,
the question ask for prime factors so for 7 it will be 1 for 8 it will be 3 as in (2^3) and 9 it will be (3^2).
i am not able to comprehend properly as the question asks number of prime factors and not the prime numbers it is constituted of..please explain..

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Re: M31-32   [#permalink] 26 Jul 2017, 08:49

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