8. For each proper divisor of $1,000,000$, take the logarithm to the base 10, add these logarithmic values together, to get the sum $S$, find the integer closest to $S$.
(Proper divisors of a natural number $n$ are the divisors of $n$ that are neither 1 nor $n$.)
8. Since $1000000=2^{\circ} \cdot 5^{\circ}$ has $(6+1)$ - $(6+1)=49$ factors (since its factors must be of the form $2^{i} \cdot 5$, where $0 \leqslant i, j \leqslant 6$), excluding 1000, the remaining 48 factors form 24 pairs, each with a product of $1000000=10^{6}$, so the product of all factors of 1000000 is $1000 \cdot\left(10^{8}\right)^{24}=10^{147}$, so the product of all proper factors of 1000000 is $10^{141}$. Since the number $S$ is the logarithm of the product of the proper factors, hence $S=141$. The integer closest to $S$ is of course 141.