2011 AMC 12B Problem 17

Problem 17 of 25IntermediateAlgebra Number Theory

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)?

Show solution

Solution

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.