The number $28!$ ($28$ in decimal) has base $30$ representation \[28!=Q6T32S??OCLQJ6000000_{30}\] where the seventh and eighth digits are missing. What are the missing digits? In base $30$, we have that the digits $A=10$, $B=11$, $C=12$, $D=13$, $E=14$, $F=15$, $G=16$, $H=17$, $I=18$, $J=19$, $K=20$, $L=21$, $M=22$, $N=23$, $O=24$, $P=25$, $Q=26$, $R=27$, $S=28$, $T=29$.

[i]2019 CCA Math Bonanza Tiebreaker Round #4[/i]
1. **Apply Wilson's Theorem**:
   - Wilson's Theorem states that for a prime \( p \), \((p-1)! \equiv -1 \pmod{p} \).
   - For \( p = 29 \), we have \( 28! \equiv -1 \pmod{29} \).
   - For \( p = 31 \), we have \( 30! \equiv -1 \pmod{31} \). Since \( 30! = 30 \times 29! \), and \( 30 \equiv -1 \pmod{31} \), it follows that \( 30! \equiv (-1) \times 29! \equiv -29! \equiv -1 \pmod{31} \). Thus, \( 29! \equiv 1 \pmod{31} \), and \( 28! \equiv 15 \pmod{31} \).

2. **Set up the equations**:
   - Let the missing digits be \( a \) and \( b \) in that order.
   - The given base 30 representation is \( 28! = Q6T32S??OCLQJ6000000_{30} \).
   - Convert the digits to their decimal equivalents: \( Q = 26 \), \( T = 29 \), \( S = 28 \), \( O = 24 \), \( C = 12 \), \( L = 21 \), \( J = 19 \).

3. **Form the congruences**:
   - For \( 28! \equiv 15 \pmod{31} \):
     \[
     6 - 19 + 26 - 21 + 12 - 24 + b - a + 28 - 2 + 3 - 29 + 9 - 26 \equiv 15 \pmod{31}
     \]
   - Simplify the left-hand side:
     \[
     -45 + b - a + 15 \equiv 15 \pmod{31}
     \]
     \[
     b - a - 30 \equiv 15 \pmod{31}
     \]
     \[
     b - a \equiv 45 \pmod{31}
     \]
     \[
     b - a \equiv 14 \pmod{31}
     \]

   - For \( 28! \equiv -1 \pmod{29} \):
     \[
     6 + 19 + 26 + 21 + 12 + 24 + b + a + 28 + 2 + 3 + 29 + 9 + 26 \equiv -1 \pmod{29}
     \]
   - Simplify the left-hand side:
     \[
     205 + b + a \equiv -1 \pmod{29}
     \]
     \[
     b + a \equiv -206 \pmod{29}
     \]
     \[
     b + a \equiv -206 + 7 \times 29 \pmod{29}
     \]
     \[
     b + a \equiv -206 + 203 \pmod{29}
     \]
     \[
     b + a \equiv -3 \pmod{29}
     \]
     \[
     b + a \equiv 26 \pmod{29}
     \]

4. **Solve the system of congruences**:
   - We have:
     \[
     b - a \equiv 14 \pmod{31}
     \]
     \[
     b + a \equiv 26 \pmod{29}
     \]
   - Adding these two equations:
     \[
     2b \equiv 40 \pmod{31}
     \]
     \[
     b \equiv 20 \pmod{31}
     \]
   - Subtracting the first equation from the second:
     \[
     2a \equiv 12 \pmod{29}
     \]
     \[
     a \equiv 6 \pmod{29}
     \]

5. **Verify the solution**:
   - \( a = 6 \) and \( b = 20 \) satisfy both congruences.
   - Therefore, the missing digits are \( a = 6 \) and \( b = 20 \).

The final answer is \( \boxed{6, 20} \).