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2011 AMC 12B Problems/Problem 17

Revision as of 23:15, 25 May 2011 by ThatDonGuy (talk | contribs)

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


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

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

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

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

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

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

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.

$h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})$, which is the 2011-digit number 8888...8889

The sum of the digits is 8 times 2010 plus 9, or $\boxed{16089\ \(\textbf{(B)}}$ (Error compiling LaTeX. Unknown error_msg)

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