2008 AMC 12B Problem 23

The sum of the base-$10$ logs of the divisors is the log of their product. A number $N$ with $d(N)$ divisors has divisor product $N^{\frac{d(N)}{2}}.$ Here $N = 10^n$ has $d(N) = (n + 1)^2$ divisors, so the product is $(10^n)^{\frac{(n+1)^2}{2}}$ and its log is $$\frac{n(n + 1)^2}{2} = 792.$$ Thus $n(n + 1)^2 = 1584$ $= 11 \cdot 144$ $= 11 \cdot 12^2,$ giving $n = 11.$ Thus, the correct answer is A.