AMC12 2008 B

AMC12 2008 B · Q23

AMC12 2008 B · Q23. It mainly tests Logarithms (rare), Counting divisors.

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

$10^n$ 的所有因子的以 $10$ 为底的对数之和为 $792$。求 $n$。

(A) 11 11

(B) 12 12

(D) 14 14

(E) 15 15

Answer

Correct choice: (A)

正确答案：（A）

Solution

Every factor of $10^n$ will be of the form $2^a \times 5^b , a\leq n , b\leq n$. Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\log(a \times b) = \log(a)+\log(b)$. For any factor $2^a \times 5^b$, there will be another factor $2^{n-a} \times 5^{n-b}$. Note this is not true if $10^n$ is a perfect square. When these are multiplied, they equal $2^{a+n-a} \times 5^{b+n-b} = 10^n$. $\log 10^n=n$ so the number of factors divided by 2 times n equals the sum of all the factors, 792. There are $n+1$ choices for the exponent of 5 in each factor, and for each of those choices, there are $n+1$ factors (each corresponding to a different exponent of 2), yielding $(n+1)^2$ total factors. $\frac{(n+1)^2}{2}*n = 792$. We then plug in answer choices and arrive at the answer $\boxed {11}$

$10^n$ 的每个因子都形如 $2^a \times 5^b$，其中 $a\leq n,\ b\leq n$。这些以 10 为底的对数并不全是有理数，但我们可以用某种方式把它们配对相加使结果为有理数。回忆对数性质 $\log(a \times b) = \log(a)+\log(b)$。对于任一因子 $2^a \times 5^b$，都存在另一个因子 $2^{n-a} \times 5^{n-b}$。注意当 $10^n$ 是完全平方数时这一点不成立。将这两个因子相乘得到 $2^{a+n-a} \times 5^{b+n-b} = 10^n$。而 $\log 10^n=n$，因此“因子个数的一半乘以 $n$”等于所有因子的对数和，即 792。每个因子中 5 的指数有 $n+1$ 种选择；对每一种选择，2 的指数也有 $n+1$ 种选择，因此共有 $(n+1)^2$ 个因子。于是 $\frac{(n+1)^2}{2}*n = 792$。代入选项可得答案 $\boxed {11}$。

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