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M31-32

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26 Jul 2017, 08:52
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.

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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26 Jul 2017, 08:55
Bunuel wrote:
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.

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.

but doesnt 8 constitute of 3 2's if we consider prime factor it would be 3 right??
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28 Nov 2017, 14:24
Beautiful Question!
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10 Mar 2018, 14:29
(I) It easy easy to fall into a trap here and think that because n= 7, 8 or 9 that this statement is insufficient. However all these numbers have only 1 prime factor.
(II) 8 has 4 factors - 8, 1, 4 and 2. Each of these numbers * each factor of n^2 will produce all the factors of 8n^2 and we are told there are 12 factors total. Therefore n^2 must have 3 factors. (3*4=12) Only perfect squares have and odd number of factors and only squares of primes have 3 factors (1, n and n^2). Therefore n is prime and has 1 prime factor
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17 Mar 2018, 03:16
For statement 2, 8n^2 has twelve factors.
It is possible in two ways,
(i) n = prime = 2 meaning n =2^4, 8n^2 is 2^3*2^8 thus 12 factors thus one prime factor only.
(i) n = prime = 5 meaning n =5, 8n^2 is 2^3*5^2 thus 12 factors but 2 different prime factors.

Implies insufficient?
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13 Jun 2018, 21:56
DTUguy wrote:
For statement 2, 8n^2 has twelve factors.
It is possible in two ways,
(i) n = prime = 2 meaning n =2^4, 8n^2 is 2^3*2^8 thus 12 factors thus one prime factor only.
(i) n = prime = 5 meaning n =5, 8n^2 is 2^3*5^2 thus 12 factors but 2 different prime factors.

Implies insufficient?

The problem is you are thinking 8n^2 as one ..
IF n=2^4 we have one prime factor . that is 2
If n= 5 , still we have one prime factor. That is 5 ( here we should not count 8 as a factor of n)
High quality question
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Re: M31-32 &nbs [#permalink] 13 Jun 2018, 21:56

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