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M31-32

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Bunuel
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M31-32 [#permalink] New post   14 Jun 2015, 14:32
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46% (01:59) correct 54% (02:24) wrong based on 65 sessions
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How many distinct prime factors does the positive integer \(n\) possess?


(1) \(44 < n^2 < 99\)

(2) \(8n^2\) has twelve factors
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Bunuel
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Re M31-32 [#permalink] New post   14 Jun 2015, 14:32
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Official Solution:


How many distinct prime factors does the positive integer \(n\) possess?

(1) \(44 < n^2 < 99\).

This statement implies that \(n\) can be 7, 8, or 9. Each of these numbers has 1 prime factor: 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 other than 2: \(8n^2 = 2^3*(prime)^2\). In this case, the number of factors can be calculated as \((3+1)(2+1) = 12\). Another possibility is when \(n = 2^4\): \(8n^2 = 2^3*(2^4)^2 = 2^{11}\). In this case, the number of factors is \(11+1 = 12\). In both cases (\(n = prime\) other than 2 or \(n = 2^4\)), \(n\) has one prime factor. Sufficient.


Answer: D
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Re: M31-32 [#permalink] New post   09 Nov 2015, 16:40
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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Bunuel
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Re: M31-32 [#permalink] New post   10 Nov 2015, 00:33
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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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Re: M31-32 [#permalink] New post   21 Jun 2016, 20:20
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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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Re: M31-32 [#permalink] New post   22 Jun 2016, 02:51
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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.

Check for more here: m31-199943.html#p1599885
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Re: M31-32 [#permalink] New post   23 Jun 2016, 05:07
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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Re: M31-32 [#permalink] New post   23 Jun 2016, 08:01
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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.

Check here: m31-199943.html#p1599885
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Re: M31-32 [#permalink] New post   17 Jul 2017, 10:26
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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Re: M31-32 [#permalink] New post   17 Jul 2017, 10:37
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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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Re: M31-32 [#permalink] New post   26 Jul 2017, 09:49
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] New post   26 Jul 2017, 09:52
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dhruv solanki 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,
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..


7 has only one prime factor, which is 7.
8 has only one prime factor, which is 2.
9 has only one prime factor, which is 3.

But for example, 36 has two primes: 2, and 3.
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Re: M31-32 [#permalink] New post   28 Nov 2017, 15:24
Beautiful Question! :thumbup:
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Re: M31-32 [#permalink] New post   16 Jun 2019, 00:20
Bunuel

I have a confusion: The question categorically asks the number of P. F and not total number of factors.
If 8n^2 has 12 factors then we can have following two possibilities:

Case 1- When n is a Prime number other than 2 - 2^3 *n^2, so we have (3+1)*(2+1)= 12 factors , in all n has 2 P.F

Case 2 - When n is not Prime and equals 16 or 2^4, so we have 2^3*2^8= 2^11= 12 factors total, in all n has 1 PF and i.e. 2 , so here we get two different answer so this statement should be insufficient
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Re: M31-32 [#permalink] New post   16 Jun 2019, 02:56
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Manku wrote:
Bunuel

I have a confusion: The question categorically asks the number of P. F and not total number of factors.
If 8n^2 has 12 factors then we can have following two possibilities:

Case 1- When n is a Prime number other than 2 - 2^3 *n^2, so we have (3+1)*(2+1)= 12 factors , in all n has 2 P.F

Case 2 - When n is not Prime and equals 16 or 2^4, so we have 2^3*2^8= 2^11= 12 factors total, in all n has 1 PF and i.e. 2 , so here we get two different answer so this statement should be insufficient


In both cases n has only 1 prime.
In case 1: n = prime other than 2 --> n has 1 prime factor.
In case 1: n = 2^4 --> n has 1 prime factor.
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Re: M31-32 [#permalink] New post   19 Jul 2020, 10:09
I think this is a high-quality question and I agree with explanation.
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Re: M31-32 [#permalink] New post   19 Jan 2022, 01:13
44<n2<99 on taking square root on both sides will result into -7 < n < 9 (approx) . Since n is positive
0<n<9. Correct me if I am wrong??
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Re: M31-32 [#permalink] New post   19 Jan 2022, 01:24
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akt715 wrote:
44<n2<99 on taking square root on both sides will result into -7 < n < 9 (approx) . Since n is positive
0<n<9. Correct me if I am wrong??


No.

Taking the square root from \(44<n^2<99\) gives: \(\sqrt{44}<|n|<\sqrt{99}\) (recall that \(\sqrt{x^2}=|x|\)), which gives: \(-9.9<n<-6.6\) or \(6.6<n<9.9\). Since n is positive, then we'd get \(6.6<n<9.9\) and since n is an integer we get n = 7, 8, or 9.
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Re: M31-32 [#permalink] New post   16 Apr 2023, 09:46
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M31-32 [#permalink] New post   04 Sep 2023, 17:49
In statement 2:
If (2^3)*(n^2) - and n is a prime number, it has 2 distinct prime numbers.
If 2^11 - there's only one distinct prime number.

How can we conclusively determine the number of distinct prime numbers?
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Re: M31-32 [#permalink]
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