Find the nth base-b digit of (b^k-1)^-2

APL (Dyalog Extended), 31 bytes

(⎕D,⎕A)⊃⍨⊃⎕⌽¯1⌽,⍉⎕(⊣⊤⍳⍤*~*-2⍨)⎕

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A full program that takes numbers from stdin in the order of k, b, n. n being 1-based cost me 3 bytes ¯1⌽.

How it works

The main mathematical trick is well explained in Neil's answer. The gist is that the repeating part of the base-b representation consists of length-k base-b digits of 0..b^k-1, except b^k-2.

This answer implements it very literally: list all repeating base-b digits, cyclically fetch nth digit from it, and convert it to 0-9A-Z. Cyclic indexing uses ngn's trick.

⎕(...)⎕     ⍝ Take k (the right) and b (the left) from stdin
  ⍳⍤*~*-2⍨  ⍝ Create a list of 0..b^k-1, and remove the number b^k-2
  ⊣⊤        ⍝ Convert these numbers into base-b digits into columns
,⍉          ⍝ Transpose and flatten to get the repeating base-b digits
⊃⎕⌽¯1⌽      ⍝ Take n from stdin and get the nth digit cyclically
(⎕D,⎕A)⊃⍨   ⍝ Convert the number to the corresponding char in 0-9A-Z