the question is asking whether k has a factor that is greater than 1, but less than itself.
if you're good at these number property rephrasings, then you can realize that this question is equivalent to "is k non-prime?", which, in turn, because it's a data sufficiency problem (and therefore we don't care whether the answer is "yes" or "no", as long as there's an answer), is equivalent to "is k prime?".
but let's stick to the first question - "does k have a factor that's between 1 and k itself?" - because that's easier to interpret, and, ironically, is easier to think about (on this particular problem) than the prime issue.
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key realization:
every one of the numbers 2, 3, 4, 5, ..., 12, 13 is a factor of 13!.
this should be clear when you think about the definition of a factorial: it's just the product of all the integers from 1 through 13. because all of those numbers are in the product, they're all factors (some of them several times over).
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consider the lowest number allowed by statement 2: 13! + 2.
note that 2 goes into 13! (as shown above), and 2 also goes into 2. therefore, 2 is a factor of this sum (answer to question prompt = "yes").
consider the next number allowed by statement 2: 13! + 3.
note that 3 goes into 13! (as shown above), and 3 also goes into 3. therefore, 3 is a factor of this sum (answer to question prompt = "yes").
etc.
all the way to 13! + 13.
works the same way each time.
so the answer is "yes" every time --> sufficient.
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