$4 \cdot 29$ Taking the logarithm to the base 10 of each proper divisor of $1,000,000$, and adding these logarithmic values, we get the sum $S$. Find the integer closest to $S$.
(4th American Invitational Mathematics Examination, 1986)
[Solution 1] Since $1,000,000=2^{6} \cdot 5^{6}$. There are $(6+1)(6+1)=49$ factors in total.

Excluding 1000, the remaining 48 factors form 24 pairs, each pair of factors multiplying to $10^{6}$, so the product of all factors of $1,000,000$ is $1000 \cdot\left(10^{6}\right)^{24}=10^{147}$.
This includes the product of the non-proper factors 1 and $10^{6}$, so the product of all proper factors is
$$
\frac{10^{147}}{10^{6}}=10^{141} \text {. }
$$

By the properties of logarithms, the logarithm of the product of these proper factors is the sum of the logarithms of each proper factor, i.e.,
$$
S=\lg 10^{141}=141 \text {. }
$$
[Solution 2] Consider the prime factors 2 and 5 in $1,000,000$ and the number of times they appear in all its factors, i.e., the power indices of 2 and 5 in the product of all factors.
Notice that $2^{j}(1 \leqslant j \leqslant 6, j \in N)$ can form the factors
$$
2^{j}, 2^{j} \cdot 5,2^{j} \cdot 5^{2}, 2^{j} \cdot 5^{3}, 2^{j} \cdot 5^{4}, 2^{j} \cdot 5^{5}, 2^{j} \cdot 5^{6} .
$$

Thus, the exponent of 2 in the product of these factors is $7 j$.
For $j$ taking $1,2,3,4,5,6$, we get
$$
7(1+2+3+4+5+6)=147 .
$$

Similarly, the exponent of 5 in the product of all factors is also 147.
Removing the exponents of 2 and 5 in the non-proper factors $1=2^{0} \cdot 5^{0}$ and $10^{6}=2^{6} \cdot 5^{6}$, the exponents of 2 and 5 in the product of all proper factors are both 141.
Thus, we have
$$
S=\lg 2^{141}+\lg 5^{141}=141
$$