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Difference between revisions of "2011 AMC 12B Problems/Problem 17"

m (Created page with '<math> \text{Let }f(x)\text{ = }10^{10x}, g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right), h_{1}(x)\text{ = }g(f(x)),\text{and }h_{n}(x)\text{ = }h_{1}(h_{n-1}(x))\\\text{…')
 
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<math> \text{Let }f(x)\text{ = }10^{10x}, g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right), h_{1}(x)\text{ = }g(f(x)),\text{and }h_{n}(x)\text{ = }h_{1}(h_{n-1}(x))\\\text{\\for integers }n\ge 2.\text{What is the sum of the digits of }h_{2011}(1)? </math>
 
<math> \text{Let }f(x)\text{ = }10^{10x}, g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right), h_{1}(x)\text{ = }g(f(x)),\text{and }h_{n}(x)\text{ = }h_{1}(h_{n-1}(x))\\\text{\\for integers }n\ge 2.\text{What is the sum of the digits of }h_{2011}(1)? </math>
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<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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<math>h_{1}(x)\text{ = }g(f(x))\text{ = }g(10^{10x})\text{ = }\text{log}_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}</math>
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Proof by induction that <math>h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math>:
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For n = 1, <math>h_{1}(x)\text{ = }10x - 1</math>
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Assume <math>h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math> is true for n:
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<math>h_{n+1}(x)\text{ = } h_{1}(h_{n}(x))\text{ = }10 h_{n}(x) - 1\text{ = 10 }(10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1
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\\= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{n-1}) - 1</math>

Revision as of 00:05, 26 May 2011

$\text{Let }f(x)\text{ = }10^{10x}, g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right), h_{1}(x)\text{ = }g(f(x)),\text{and }h_{n}(x)\text{ = }h_{1}(h_{n-1}(x))\\\text{\\for integers }n\ge 2.\text{What is the sum of the digits of }h_{2011}(1)?$


$g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right)\text{ = }\text{log}_{10}\left({x}\right)\text{ - 1}$

$h_{1}(x)\text{ = }g(f(x))\text{ = }g(10^{10x})\text{ = }\text{log}_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}$

Proof by induction that $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$:

For n = 1, $h_{1}(x)\text{ = }10x - 1$

Assume $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ is true for n:

$h_{n+1}(x)\text{ = } h_{1}(h_{n}(x))\text{ = }10 h_{n}(x) - 1\text{ = 10 }(10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1 \\= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{n-1}) - 1$

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