Question

$10^n$ 的所有正因数的以 $10$ 为底的对数之和为 $792$ 。求 $n$ ？

The sum of the base- $10$ logarithms of the divisors of $10^n$ is $792.$ What is $n?$

$11$

$12$

$13$

$14$

$15$

所有因数的以 $10$ 为底的对数之和，就是这些因数乘积的对数。若一个数 $N$ 有 $d(N)$ 个因数，则其所有因数的乘积为 $N^{d(N)/2}$ 。

这里 $N = 10^n$ 有 $d(N) = (n + 1)^2$ 个因数，所以因数乘积为 $(10^n)^{(n+1)^2/2}$ ，其对数为 $\frac{n(n + 1)^2}{2} = 792.$

因此 $n(n + 1)^2 = 1584$ $= 11 \cdot 144$ $= 11 \cdot 12^2$ ，得到 $n = 11$ 。

因此，正确答案是 A。

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 $N^{d(N)/2}.$

Here $N = 10^n$ has $d(N) = (n + 1)^2$ divisors, so the product is $(10^n)^{(n+1)^2/2}$ 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.