Difference between revisions of "2011 AMC 12B Problems/Problem 17"
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<math>g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1</math> | <math>g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1</math> | ||
− | <math>h_{1}(x)=g(f(x))\text{ = }g(10^{10x}=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}</math> | + | <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>: |
Latest revision as of 15:40, 21 September 2024
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
Solution 2 (Quick, Non-Rigorous Trends)
As before, . Compute , , and to yield 9, 89, and 889. Notice how this trend will repeat this trend (multiply by 10, subtract 1, repeat). As such, is just 2010 8's followed by a nine. .
~~BJHHar
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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