Problem 1284

1. **Define $A$:**
Let $A$ be the number of 2019-digit numbers made of 2 different digits. We can split $A$ into two parts:
$$A = A_0 + A_1$$
where $A_0$ is the number of 2019-digit numbers made of the digit $0$ and another digit $k$ (where $1 \le k \le 9$), and $A_1$ is the number of 2019-digit numbers made of two different non-zero digits.

2. **Calculate $A_0$:**
- A number containing only the digits $k$ and $0$ must begin with $k$ (since it is a 2019-digit number and cannot start with 0).
- The remaining 2018 digits can be any combination of $k$ and $0$, except the configuration with all 2018 digits being $k$.
- For a fixed $k$, there are $2^{2018} - 1$ such numbers.
- Since there are 9 possible values for $k$ (from 1 to 9), we have:
$$A_0 = 9 \times (2^{2018} - 1)$$

3. **Calculate $A_1$:**
- Let $a, b \in \{1, 2, \ldots, 9\}$ with $a \neq b$.
- A number which begins with the digit $a$ can continue with any combination of $a$ and $b$, except the configuration with all 2018 digits being $a$.
- Similarly, a number which begins with the digit $b$ can continue with any combination of $a$ and $b$, except the configuration with all 2018 digits being $b$.
- Hence, there are $2 \times (2^{2018} - 1)$ numbers containing only the digits $a$ and $b$ (i.e., $2^{2018} - 1$ numbers beginning with $a$ and $2^{2018} - 1$ numbers beginning with $b$).
- With the elements of the set $\{1, 2, \ldots, 9\}$, we can form $\binom{9}{2} = 36$ distinct pairs $(a, b)$ (since $(a, b)$ and $(b, a)$ are not distinct).
- Therefore:
$$A_1 = 36 \times 2 \times (2^{2018} - 1) = 72 \times (2^{2018} - 1)$$

4. **Combine $A_0$ and $A_1$:**
$$A = A_0 + A_1 = 9 \times (2^{2018} - 1) + 72 \times (2^{2018} - 1) = 81 \times (2^{2018} - 1)$$

5. **Determine the highest power of 3 that divides $A$:**
$$A = 81 \times (2^{2018} - 1) = 3^4 \times (2^{2018} - 1)$$
We need to find $\nu_3(2^{2018} - 1)$.

6. **Calculate $\nu_3(2^{2018} - 1)$:**
- We use the fact that $2^6 \equiv 1 \pmod{9}$ (since $2^6 = 64 \equiv 1 \pmod{9}$).
- Since $2018 = 6 \times 336 + 2$, we have:
$$2^{2018} = (2^6)^{336} \times 2^2 \equiv 1^{336} \times 4 \equiv 4 \pmod{9}$$
- Therefore:
$$2^{2018} - 1 \equiv 4 - 1 \equiv 3 \pmod{9}$$
- This implies:
$$\nu_3(2^{2018} - 1) = 1$$

7. Combine the results:
$$\nu_3(A) = \nu_3(3^4) + \nu_3(2^{2018} - 1) = 4 + 1 = 5$$

The final answer is $\boxed{ \nu_3(A) = 5 }$