Take a moment to learn the following property; it will provide a useful way to think of divisibility questions.

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!---------------------

30! = (30)(29)(28)(27).......(4)(3)(2)(1)

Let's check each answer choice....
A) 58
58 = (29)(2)
Since 30! = (30)(29)(28)(27).......(4)(3)(2)(1), we can conclude that 30! is divisible by 58
Eliminate answer choice A


B) 62
62 = (2)(31)
Since there are no 31's hiding in the prime factorization of 30!, we can conclude that 30! is NOT divisible by 62

Answer: B

For "fun," let's show that 30! is not divisible by each of the remaining three answer choices...

C) 66
66 = (2)(3)(11)
Since 30! = (30)(29)(28)(27)....... (11)(10)(9)...(4)(3)(2)(1), we can conclude that 30! is divisible by 66
Eliminate answer choice C

D) 68
68 = (2)(2)(17)
Since 30! = (30)(29)(28)(27)....... (17)...(2)(2)(3)(2)(1), we can conclude that 30! is divisible by 68
Eliminate answer choice D

E) 72 = (18)(2)(2)
Since 30! = (30)(29)(28)(27)....... (18)...(2)(2)(3)(2)(1), we can conclude that 30! is divisible by 72
Eliminate answer choice E