Any number of the form p^{a}q^{b}r^{c} will have (a + 1) (b + 1) (c + 1) factors, where p, q, r are prime. In order for the number to be a perfect cube a, b, c will have to be multiples of 3.
We can assume that a = 3m, b = 3n, c = 3l.
This tells us that the number of factors will have to be of the form (3n + 1) * (3m + 1) * (3l + 1). In other words (a + 1), (b + 1) and (c + 1) all leave a remainder of 1 on division by 3. So, the product of these three numbers should also leave a remainder of 1 on division by 3. Of the four numbers provided, 16 and 28 can be written in this form, the other two cannot..
So, a perfect cube can have 16 or 28 factors. Now, let us think about what kind of numbers will have 16 factors..
A number of the form p^{15} or q^{3}r^{3} will have exactly 16 factors. Both are perfect cubes. Note that there are other prime factorizations possible that can have exactly 16 factors. But these two forms are perfect cubes, which is what we are interested in.
Similarly, a number of the form p^{27} or q^{3}r^{6} will have 28 factors. Both are perfect cubes.
Correct Answer: A and B
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