Hello, everyone:
Great thread - it looks like you're all on board with why the correct answer here is B, but if you don't mind I'd also like to point out why I had an impossible time not butting in here!
If you can think about any factorial, it's divisible by every number within it, pretty much by definition. Let's use 4! as an example:
4*3*2*1 is divisible by 4, 3, 2, and 1
Now, when you take that factorial and add 1, it can no longer be divisible by any of the prime factors (or, really, factors >1) that it previously was:
4*3*2*1 = 24, add 1 and it's 25, which is not divisible by 4, 3, or 2
The reason behind that is that:
Every second number (2, 4, 6, 8, 10...) is divisible by 2
Every third number (3, 6, 9, 12...) is divisible by 3
Every fourth number (4, 8, 12, 16) is divisible by 4
and so on...
If you take factorial and add 1, it's off of each of those cycles. You'd need to add 2 to keep it even, add 3 to keep it divisible by 3, etc. Only adding 1 just takes it off of each cycle that it was originally on.
So, for this problem, if you add any number between 2 and 17, you'll keep it on one of its previous factor cycles: add 3 and it's still divisible by 3; add 5 and it's still divisible by 5, etc.
If you had added 1, then there's a chance that it would be prime, but the number is so huge that you wouldn't really be able to calculate it as it might catch another, larger prime factor somehow. If they wanted to make it really difficult, though, they could have added 18 as a possible additive term (17! + 18). That would not be prime...17! is divisible by 2, so adding any even number to it would keep it even. You'd need to add 1 or a prime number greater than 17 in order to have the possibility of a prime number.
The more you think about problems like these, the better equipped you'll be to think quickly about them on test day. Thanks for bringing this one up and keeping the thread alive!
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
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