Problem:
Let f(x)=1010x,g(x)=log10​(10x​),h1​(x)=g(f(x)), and hn​(x)=h1​(hn−1​(x)) for integers n≥2. What is the sum of the digits of h2011​(1) ?
Answer Choices:
A. 16,081
B. 16,089
C. 18,089
D. 18,098
E. 18,099
Solution:
Note that
h1​(x)=log10​(101010x​)=log10​(1010x−1)=10x−1
Therefore h2​(x)=102x−(1+10),h3​(x)=103x−(1+10+102), and in general,
hn​(x)=10nx−k=0∑n−1​10k
Hence hn​(1) is an n-digit integer whose units digit is 9 and whose other digits are all 8 's. The sum of the digits of h2011​(1) is 8⋅2010+9=16,089​.
The problems on this page are the property of the MAA's American Mathematics Competitions