Difference between revisions of "2011 AMC 12B Problems/Problem 17"
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<math>h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})</math>, which is the 2011-digit number 8888...8889 | <math>h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})</math>, which is the 2011-digit number 8888...8889 | ||
| − | 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 == | == See also == | ||
{{AMC12 box|year=2011|num-b=16|num-a=18|ab=B}} | {{AMC12 box|year=2011|num-b=16|num-a=18|ab=B}} | ||
Revision as of 21:19, 26 September 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 | |
