AMC12 2011 B

AMC12 2011 B · Q17

AMC12 2011 B · Q17. It mainly tests Exponents & radicals, Logarithms (rare).

Let $f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))$, and $h_n(x) = h_1(h_{n-1}(x))$ for integers $n \geq 2$. What is the sum of the digits of $h_{2011}(1)$?

设 $f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))$，且对整数 $n \geq 2$，$h_n(x) = h_1(h_{n-1}(x))$。$h_{2011}(1)$ 的各位数字之和是多少？

(A) 16081 16081

(B) 16089 16089

(D) 18098 18098

(E) 18099 18099

Answer

Correct choice: (B)

正确答案：（B）

Solution

$g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1$ $h_{1}(x)=g(f(x))\text{ = }g(10^{10x})=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}$ Proof by induction that $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$: For $n=1$, $h_{1}(x)=10x - 1$ Assume $h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ is true for n: \begin{align*} h_{n+1}(x)&= h_{1}(h_{n}(x))\\ &=10 h_{n}(x) - 1\\ &=10 (10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1\\ &= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1\\ &= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) \end{align*} Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n. $h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})$, which is the 2011-digit number 8888...8889 The sum of the digits is 8 times 2010 plus 9, or $\boxed{16089\textbf{(B)}}$

$g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1$ $h_{1}(x)=g(f(x))\text{ = }g(10^{10x})=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}$ 用归纳法证明 $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$：当 $n=1$ 时，$h_{1}(x)=10x - 1$ 假设对 $n$，$h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ 成立： \begin{align*} h_{n+1}(x)&= h_{1}(h_{n}(x))\\ &=10 h_{n}(x) - 1\\ &=10 (10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1\\ &= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1\\ &= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) \end{align*} 因此，若对 $n$ 成立则对 $n+1$ 也成立；又因对 $n = 1$ 成立，所以对所有正整数 $n$ 都成立。 $h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})$，这是一个 2011 位数 8888...8889。各位数字之和为 $8$ 乘以 $2010$ 再加 $9$，即 $\boxed{16089\textbf{(B)}}$。

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