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

Let's say the integer in question is N

The integer (N) is divisible by 27
27 = (3)(3)(3)
So, we can say N = (3)(3)(3)(?)(?)...

Aside: the other ?'s represents other possible prime numbers in the prime factorization of N

The integer (N) is divisible by 10
10 = (2)(5)
So, we can say N = (2)(5)(?)(?)...

When we combine both pieces of information we can write: N = (3)(3)(3)(2)(5)(?)(?)

The integer (N) also must be divisible by which of the following?
N = (3)(3)(3)(2)(5)(?)(?)

A. 4
4 = (2)(2)
Since the prime factorization of N (above) does NOT include two 2's, it need not be the case that N is divisible by 4
Eliminate A

B. 25
25 = (5)(5)
Since the prime factorization of N (above) does NOT include two 5's, it need not be the case that N is divisible by 25
Eliminate B

C. 36
36 = (2)(2)(3)(3)
Since the prime factorization of N (above) does NOT include two 2's and two 3's, it need not be the case that N is divisible by 36
Eliminate C

D. 54
54 = (2)(3)(3)(3)
Since the prime factorization of N (above) DOES include one 2 and three 3's, we can be certain that N is divisible by 54


Answer: D