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−110k
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