Let f(x)=10^10x, g(x)=log_10 ((x)/(10)), h_1(x)=g(f(x)), and h_n(x)=h_1(h_n-1(x)) for integers n≥2. What is the sum of the digits of h_2011(1)?

2011 AMC 12B Problem 17

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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

Simplify

h_1(x)=\log_{10}\!\left(\dfrac{10^{10x}}{10}\right)

to a linear function

Iterating gives

h_n(1)

as a number whose digits are all

8

’s ending in a

9

First,

\begin{aligned} h_1(x) &=\log_{10}\!\left(\dfrac{10^{10x}}{10}\right) \\ &=\log_{10}\!\left(10^{10x-1}\right) \\ &=10x-1. \end{aligned}

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.