Problem 1107

1. First, we need to determine the proper divisors of $1000000$. We start by expressing $1000000$ in its prime factorized form:
$$1000000 = 10^6 = (2 \times 5)^6 = 2^6 \times 5^6$$
The number of divisors of $1000000$ is given by $(6+1)(6+1) = 49$. Since we are interested in proper divisors, we exclude $1000000$ itself, leaving us with $48$ proper divisors.

2. For any proper divisor $a$ of $1000000$, there exists another proper divisor $b$ such that $ab = 1000000$. This means:
$$\log_{10} a + \log_{10} b = \log_{10} (a \times b) = \log_{10} 1000000 = 6$$
Therefore, each pair of proper divisors $(a, b)$ contributes $6$ to the sum of the logarithms.

3. Since there are $48$ proper divisors, they can be paired into $24$ pairs. Each pair contributes $6$ to the sum, so the total contribution from these pairs is:
$$24 \times 6 = 144$$

4. However, we must consider the divisor $1000$ (which is $10^3$). This divisor pairs with itself, contributing:
$$\log_{10} 1000 = 3$$
Since $1000$ is a proper divisor, it should be included in the sum.

5. Therefore, the total sum $S$ of the base 10 logarithms of all the proper divisors of $1000000$ is:
$$S = 144 - 3 = 141$$

The final answer is $\boxed{141}$.