aharonm wrote:

Thanks!

So i guess if st2 was one of the following : n=3 or n=4 or n=5 we would still get B as the right answer

cause the only opition a=2

but if st2 was n=2 it would have been E cause a=2 or a=3

right?

That's not quite right:

8! = (2^7)(3^2)(5^1)(7^1)

If n=6, 5 or 4, then a must be 2. If, however, n=3, then we have two possible values for a. a could again certainly be equal to 2, but a could also be equal to 4, since 4^3 = (2^2)^3 = 2^6, which is a divisor of 8!. If n = 2, there are quite a few possibilities: a could be 2, 3, 4, 6, 8, 12 or 24.

Of course as soon as you combine the statements, knowing that a is positive, you can definitely solve for a without even reading the rest of the information in the question. That's an immediate clue that C won't likely be the answer here - why would they bother including the information about 8! if it wasn't important in the solution of the problem?

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