1. A number is divisible by $6$ if and only if it is divisible by both $2$ and $3$.

2. The six-digit numbers are formed using only the digits $4,5,9$.

3. Condition for divisibility by $2$: The last digit must be even. Among $4,5,9$, only $4$ is even.

   So, the last digit is fixed as $4$.

4. Now the first $5$ places can each be filled independently using $4,5,9$. Hence total such six-digit numbers ending in $4$: $3^5 = 243$

5. Condition for divisibility by $3$: Sum of digits must be divisible by $3$.

   Now, $4 \equiv 1 \pmod{3}, \quad 5 \equiv 2 \pmod{3}, \quad 9 \equiv 0 \pmod{3}$

   Since the last digit is $4$, it contributes $1 \pmod{3}$. Therefore, the sum of the first five digits must be congruent to $2 \pmod{3}$.

6. Let us count 5-digit strings from $\{4,5,9\}$ whose digit-sum is $2 \pmod{3}$.

   Each position contributes residue $1,2,0$ respectively, and each choice is equally available.

   For strings of length $5$, the residues modulo $3$ are equally distributed among $0,1,2$ because:

   • total number of strings is $3^5 = 243$,
   • by symmetry (or roots of unity filter), each residue class occurs equally often.

   Hence the number of 5-digit strings with sum congruent to $2 \pmod{3}$ is $\frac{3^5}{3} = 81$

7. Therefore, the total number of six-digit numbers divisible by $6$ is $81$

8. Comparison with stored answer: Stored correct answer = $81$. Our derived answer also = $81$.

So the stored answer is correct.