Difference between revisions of "2011 AMC 12B Problems/Problem 17"
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==Solution== | ==Solution== | ||
− | <math>g(x) | + | <math>g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1</math> |
− | <math>h_{1}(x) | + | <math>h_{1}(x)=g(f(x))\text{ = }g(10^{10x}=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}</math> |
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> | + | For <math>n=1</math>, h_{1}(x)=10x - 1<math> |
− | Assume <math>h_{n}(x) | + | Assume </math>h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})<math> is true for n: |
− | + | \begin{align*} | |
− | \= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1 | + | h_{n+1}(x)&= h_{1}(h_{n}(x))\\ |
− | \= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) | + | &=10 h_{n}(x) - 1\\ |
+ | &=10 (10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1\ | ||
+ | &= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1\ | ||
+ | &= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) | ||
+ | \end{align*} | ||
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>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)}} | + | The sum of the digits is 8 times 2010 plus 9, or </math>\boxed{16089\textbf{(B)}}$ |
== 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 11:29, 11 October 2018
Problem
Let , and for integers . What is the sum of the digits of ?
Solution
Proof by induction that :
For , h_{1}(x)=10x - 1h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$is true for n:
Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n.$ (Error compiling LaTeX. Unknown error_msg)h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})$, which is the 2011-digit number 8888...8889
The sum of the digits is 8 times 2010 plus 9, or$ (Error compiling LaTeX. Unknown error_msg)\boxed{16089\textbf{(B)}}$
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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