-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----ONTO THE QUESTION!---------------------


9! = (9)(8)(7)(6)(5)(4)(3)(2)(1)
So, 1 to 9 are definitely factors of 9!
10 is also a factor of 9! since 9! = (9)(8)(7)(6)(5)(4)(3)(2)(1) = (9)(8)(7)(6)(10)(4)(3)(1)
11 is prime, so we can ignore that.
12 is also a factor of 9! since 9! = (9)(8)(7)(6)(5)(4)(3)(2)(1) = (9)(8)(7)(6)(12)(4)(3)(1)

Using the same logic, we can show that 14, 15, 16, 18, 20 and 21 are all factors of 9!

However, 22 is NOT a factor of 9!
We know this because 22 = (2)(11) and there is no 11 hiding in the prime factorization of 9!

Answer: 22

Cheers,
Brent