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(1) \(15!<z\) --> \(z\) is more than some number (\(15!\)). \(z\) may or may not be a prime. Not sufficient.

(2) \(17!+2\leq{z}\leq{17!+17}\) --> \(z\) cannot be a prime. For instance if \(z=17!+13=13*(2*4*5*6*7*8*9*10*11*12*14*15*16*17+1)\), then \(z\) is a multiple of 13, so not a prime. Same for all other numbers in this range. So, \(z=17!+x\), where \(2\leq{x}\leq{17}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(17!+x\), the same way as we did for 13). Sufficient.

For the above question, this will help (its a matched question and all the posts there make things clearer) if-x-is-an-integer-does-x-have-a-factor-n-such-that regards

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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