Paraphrasing an earlier Reddit post, here's how it might look in K3.

Generate a range of numbers up to 20 (for brevity), inclusive:

      1+!20
    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Apply a function ({x!/:3 5}), which takes an element modulo 3 and 5, to the range:

      {x!/:3 5}1+!20
    (1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2
     1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0)

Taking the negation (~) shows us the places the elements divide evenly:

      {~x!/:3 5}1+!20
    (0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0
     0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1)

Convert these into base-2 indices:

      {2 _sv~x!/:3 5}1+!20
    0 0 2 0 1 2 0 0 2 1 0 2 0 0 3 0 0 2 0 1

We can then use those indices to select from a list. Note the addition of a map ('), because if we continue performing all these operations in parallel we won't have access to x as some particular item of the list:

      {(x;"Buzz";"Fizz";"FizzBuzz")2 _sv~x!/:3 5}'1+!20
    (1;2;"Fizz";4;"Buzz";"Fizz";7;8;"Fizz";"Buzz";11;"Fizz";13;14;"FizzBuzz";16;17;"Fizz";19;"Buzz")