I'm implicitly using the method to count divisors from a prime factorization below - if that method is unfamiliar to anyone reading, this solution will not make sense:

Using Statement 1 alone, if n is not divisible by 3, then the only way (3^3)(n^3) can have 16 factors is if n is prime. So n might be a prime number, and have only two divisors.

But there's another possibility: n might be a power of 3. If n = 3^4, then (3^3)(n^3) = (3^3)(3^12) = 3^15, which also has 16 divisors. So n might equal 3^4, which has five divisors, and Statement 1 is not sufficient.

Statement 2 tells us n = 5 so it is sufficient, and the answer is B.
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(1) 3^3 * n^3 has 16 factors.
-> n is a prime number. Prime number has only two factors which is 1 and itself.
SUFFICIENT

(2) Only n=5 is possible.
SUFFICIENT

IMO. (D)

Asked: How many divisors does the positive integer n have?


(1) \(27n^3\) has 16 factors.
3^3*n^3 has 16 factors
(3+1)(3+1) = 16
n is a prime number
n has 2 divisors
SUFFICIENT

(2) \(90 < n^3 < 200\)
n=5; n^3 = 125
n has 2 divisors
SUFFICIENT

IMO D



[/quote]
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I have neglected the fact that n can be 3^4, not only prime number.
Thank you so much for your explanation!

Posted from my mobile device

f(n:posinteger)=?

(1) insufic
f(3^3n^3)=16
n=prime!=3: (3+1)(3+1)=16, n has 2 factors
n=3^4: 3^3(3^12)=3^15=16 factors, n has 5 factors

(2) sufic
4^3=64, 5^3=125, 6^3=216
n=5

(B)

(1) 27n327n3 has 16 factors.
3^3*n^3 has 16 factors
(3+1)(X) factors = 16
n must be a prime number => n has 2 divisors
SUFFICIENT

(2) 90<n3<20090<n3<200
Only n=5 is possible in this range; n^3 = 125
n has 2 divisors
SUFFICIENT

=> Both are sufficient!
IMO D

IanStewart
Nice! missed the other possibility. How did you think so? Would be great if you could please share your insights on the thought process here.

If you know, for example, that the integer n has 100 positive divisors, and you want to imagine what the prime factorization of n might look like, there are lots of possibilities -- it might look like p^4 * q^4 * r * s, say, or like p^49 * q. But the simplest possibility is that the prime factorization looks like p^99. Since we very often want to come up with the simplest possible examples when trying to prove a DS statement is not sufficient, in questions like the one in this thread, the example that occurs to me first is that the number might be 3^15. It then became a question of whether there was a second possible scenario, and there is, 3^3 * p^3.

I'd add though that it's not likely to be important on the actual GMAT to know how to think up simple numbers when you're told "positive integer k has 12 divisors". It seems there's dozens upon dozens of questions on this forum, and in prep company books, about counting divisors. Questions about that do occasionally appear on the GMAT, but they are rare, and they seem to almost never be as complicated as most company questions about the subject.
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Statement 1 : 27*n^3 = 3^3 *n^3; now this has 4*x =16 factors that means n^3 has total 4 factors. These 4 factors could be same numbers or different numbers -> hence not sufficient.

Statement 2: 90< n^3 <200 -> only 5 satisfies the condition -> sufficient

Hence answer is B

IanStewart - Thanks a lot!

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