Let $S$ be the sum of the base 10 logarithms of all the proper divisors of 1000000. What is the integer nearest to $S$?
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 \( \boxed{141} \).