2011 AMC 12B Problem 17

Below is the professionally curated solution for Problem 17 of the 2011 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12B solutions, or check the answer key.

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Concepts:logarithm recursion digits

Difficulty rating: 1980

17.

Let $f(x)=10^{10x},$ $g(x)=\log_{10}\!\left(\dfrac{x}{10}\right),$ $h_1(x)=g(f(x)),$ and $h_n(x)=h_1(h_{n-1}(x))$ for integers $n\ge2.$ What is the sum of the digits of $h_{2011}(1)?$

$16{,}081$

$16{,}089$

$18{,}089$

$18{,}098$

$18{,}099$

Solution:

First, $h_1(x)=\log_{10}\!\left(\dfrac{10^{10x}}{10}\right)=\log_{10}\!\left(10^{10x-1}\right)=10x-1.$

Iterating, $h_n(x)=10^n x-(1+10+\cdots+10^{n-1}).$ Therefore $h_n(1)$ is an $n$ -digit integer whose units digit is $9$ and all of whose other digits are $8.$

For $n=2011,$ the digit sum is $8\cdot2010+9=16{,}089.$

Thus, the correct answer is B.

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