Difference between revisions of "2011 AMC 12B Problems/Problem 17"
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Proof by induction that <math>h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math>: | Proof by induction that <math>h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math>: | ||
− | For <math>n=1</math>, h_{1}(x)=10x - 1<math> | + | For <math>n=1</math>, <math>h_{1}(x)=10x - 1</math> |
− | Assume < | + | Assume <math>h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math> is true for n: |
\begin{align*} | \begin{align*} | ||
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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. | 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. | ||
− | < | + | <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 < | + | 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:30, 11 October 2018
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
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 |
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