Problem 1335

1. Understanding the Problem:
We need to find the probability $q$ that the determinant of a randomly chosen $n \times n$ matrix $A$ with entries from $\{0, 1, \ldots, p-1\}$ is congruent to 1 modulo $p$ as $n \to \infty$. Then, we need to compute the sum of the first $p^2$ digits after the decimal point in the base $p$ expansion of $q$ and find the remainder when this sum is divided by $10^9$.

2. Probability Calculation:
Let $S_i$ be the set of $n \times n$ matrices with determinant $i \mod p$. Since the determinant function is multiplicative and the field $\mathbb{Z}/p\mathbb{Z}$ is finite, the sizes of these sets are equal for $i \in \{1, 2, \ldots, p-1\}$. The number of non-singular matrices is given by:
$$P = \prod_{i=0}^{n-1} (p^n - p^i)$$
The probability $q_n$ that $\det A \equiv 1 \pmod{p}$ is:
$$q_n = \frac{1}{p-1} \cdot \frac{P}{p^{n^2}}$$
As $n \to \infty$, this probability converges to a limit $q$.

3. **Base $p$ Expansion of $q$:**
The base $p$ expansion of $q$ is given by:
$$q = 0.09889001111099988888890000111111110999998888\ldots_p$$
The significant transitions occur at pentagonal numbers. Let $f(k) = \frac{k(3k-1)}{2}$ and $g(k) = \frac{k(3k+1)}{2}$.

4. Summing the Digits:
For odd $k$, the digits from $f(k)+1$ to $f(k+1)$ contribute $2k$ since:
$$[f(k)+1, g(k)]: 0 \quad \text{and} \quad [g(k)+1, f(k+1)-1]: 1 \quad \text{and} \quad f(k+1): 0$$
For even $k$, the digits from $f(k)$ to $f(k+1)$ contribute $6046k + 2016$ since:
$$[f(k)+1, g(k)]: 9 \quad \text{and} \quad [g(k)+1, f(k+1)-1]: 8 \quad \text{and} \quad f(k+1): 9$$

5. Calculating the Sum:
Since $f(1647) < 2017^2 < g(1647)$, we consider up to $k = 1646$:
- For odd $k$ (1, 3, ..., 1645), we get:
$$6046 \left(1 + 3 + \ldots + 1645\right) + 2016 \cdot 823 = 6046 \cdot 823^2 + 2016 \cdot 823$$
- For even $k$ (2, 4, ..., 1646), we get:
$$2 \left(2 + 4 + \ldots + 1646\right) = 4 \cdot 823 \cdot \frac{823 + 1}{2}$$
- Since $2017^2 - f(1647)$ contains only the digit 2016, we get:
$$[2017^2 - f(1647)] \cdot 2016$$

6. Summing All Contributions:
Summing all contributions, we get:
$$6046 \cdot 823^2 + 2016 \cdot 823 + 4 \cdot 823 \cdot 412 + (2017^2 - f(1647)) \cdot 2016$$
Simplifying, we find the sum to be $4,098,547,790$.