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# M31-32

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Math Expert
Joined: 02 Sep 2009
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14 Jun 2015, 14:32
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95% (hard)

Question Stats:

48% (01:18) correct 52% (01:46) wrong based on 63 sessions

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How many different prime factors does positive integer $$n$$ have?

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

(2) $$8n^2$$ has twelve factors

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14 Jun 2015, 14:32
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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.

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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.

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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10 Nov 2015, 00:33
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.

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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21 Jun 2016, 20:20
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?

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22 Jun 2016, 02:51
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?

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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23 Jun 2016, 04:21
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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23 Jun 2016, 04:27
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sagarnamo1 wrote:
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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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

Thanks![quote][/quote]
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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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23 Jun 2016, 08:01
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.

Check here: m31-199943.html#p1599885
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24 Jun 2016, 00:40
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.

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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27 Jun 2016, 09:34
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.

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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09 Jul 2016, 04:15
I think this is a high-quality question and I agree with explanation.
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11 Oct 2016, 14:27
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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11 Oct 2016, 21:56
monish447 wrote:
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.

I think you mean that n can be 2^4. Still in this case n would have only one prime.
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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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17 Jul 2017, 10:37
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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17 Jul 2017, 18:18
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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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.

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..
Re: M31-32 &nbs [#permalink] 26 Jul 2017, 09:49

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# M31-32

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