2008 AMC 12B 第 23 题

The sum of the logs of the divisors equals the log of the product of all divisors

$10^n$ has $(n + 1)^2$ divisors, and the product of the divisors of $N$ is $N^{\frac{d(N)}{2}};$ this gives $\tfrac12 n(n + 1)^2 = 792$

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.