-----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 a multiple of k, then k is "hiding" within the prime factorization of N

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

Given: X is a multiple of 8
Since 8 = (2)(2)(2), we know that there are three 2's hiding in the prime factorization of X
So, X = (2)(2)(2)(?)(?)(?)....
Note: The additional (?)'s represents other possible prime numbers in the prime factorization of X.
For the moment, however, all we know for certain is that there are three 2's hiding in the prime factorization of X

Given: X is a multiple of 14
14 = (2)(7), we know that there is one 2 and one 7 hiding in the prime factorization of X
Since we already have a 2 in our prime factorization above, we need only add a 7 to our prime factorization.
We get: X = (2)(2)(2)(7)(?)(?)(?)....

Given: X is a multiple of 33
Since 33 = (3)(11), we know that there is one 3 and one 11 hiding in the prime factorization of X
So let's add them to our prime factorization.
We get: X = (2)(2)(2)(7)(3)(11)(?)(?)(?)....

So, X = (2)(2)(2)(3)(7)(11)(?)(?)(?)....
We can now see that X is a multiple of 8, 14, and 33

Now let's turn our attention to the answer choices:
A. 16
16 = (2)(2)(2)(2)
So, in order for 16 to be a factor, X must contain four 2's in its prime factorization
Since X does NOT contain four 2's in its prime factorization, 16 is NOT a factor of X

B. 24
24 = (2)(2)(2)(3)
Since X = (2)(2)(2)(3)(7)(11), we can see that 24 IS a factor of X

C. 77
77 = (7)(11)
Since X = (2)(2)(2)(3)(7)(11), we can see that 77 IS a factor of X

D. 81
81 = (3)(3)(3)(3)
So, in order for 81 to be a factor, X must contain four 3's in its prime factorization
Since X does NOT contain four 3's in its prime factorization, 81 is NOT a factor of X


Answer: B, C

Cheers,
Brent