Difference between revisions of "2011 AMC 12B Problems/Problem 17"
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| − | <math> | + | ==Problem== |
| + | Let <math>f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))</math>, and <math>h_n(x) = h_1(h_{n-1}(x))</math> for integers <math>n \geq 2</math>. What is the sum of the digits of <math>h_{2011}(1)</math>? | ||
| + | <math>\textbf{(A)}\ 16081 \qquad \textbf{(B)}\ 16089 \qquad \textbf{(C)}\ 18089 \qquad \textbf{(D)}\ 18098 \qquad \textbf{(E)}\ 18099</math> | ||
| + | |||
| + | ==Solution== | ||
<math>g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right)\text{ = }\text{log}_{10}\left({x}\right)\text{ - 1}</math> | <math>g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right)\text{ = }\text{log}_{10}\left({x}\right)\text{ - 1}</math> | ||
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The sum of the digits is 8 times 2010 plus 9, or <math>\boxed{16089\ \(\textbf{(B)}}</math> | The sum of the digits is 8 times 2010 plus 9, or <math>\boxed{16089\ \(\textbf{(B)}}</math> | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2011|num-b=16|num-a=18|ab=B}} | ||
Revision as of 17:23, 29 May 2011
Problem
Let 
, and 
for integers 
. What is the sum of the digits of 
?
Solution
Proof by induction that 
:
For 
Assume is true for n:
Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n.
, which is the 2011-digit number 8888...8889
The sum of the digits is 8 times 2010 plus 9, or $\boxed{16089\ \(\textbf{(B)}}$ (Error compiling LaTeX. Unknown error_msg)
See also
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
