Let's find the prime factorization of the numerator and denominator and then simplify.

We get: \(\frac{26k}{84}=\frac{(2)(13)k}{(2)(2)(3)(7)}=\frac{(13)k}{(2)(3)(7)}\)

In order for \(\frac{(13)k}{(2)(3)(7)}\) to be an INTEGER, k must be divisible by 2, 3 and 7.

(2)(3)(7) = 42, which means k must be divisible by 42.
Put another way, k must be a multiple of 42 (as long as k is less than 215, as stated in the question)

If k = 42, then the prime factors of k are 2, 3 and 7
If k = (2)(42) (a multiple of 42), then the prime factors of k are 2, 3 and 7
If k = (3)(42) (a multiple of 42), then the prime factors of k are 2, 3 and 7
If k = (4)(42) (a multiple of 42), then the prime factors of k are 2, 3 and 7
If k = (5)(42) (a multiple of 42), then the prime factors of k are 2, 3, 5 and 7

So, k can have, AT MOST, four different prime factors

Answer: C