2011 AMC 12B Problems/Problem 17: Difference between revisions
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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> | ||
<math>g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right)\text{ = }\text{log}_{10}\left({x}\right)\text{ - 1}</math> | |||
<math>h_{1}(x)\text{ = }g(f(x))\text{ = }g(10^{10x})\text{ = }\text{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>: | |||
For n = 1, <math>h_{1}(x)\text{ = }10x - 1</math> | |||
Assume <math>h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math> is true for n: | |||
<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 | |||
\\= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{n-1}) - 1</math> | |||
Revision as of 23:05, 25 May 2011
Proof by induction that 
:
For n = 1, 
Assume is true for n:
