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2005 AMC 12A Problems/Problem 20

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Problem

For each in , define \begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\f(x) & = 2 - 2x, & \text { if } \frac {1}{2} < x \leq 1. \end{array} Let , and for each integer . For how many values of in is ? (\text {A}) \ 0 \qquad (\text {B}) \ 2005 \qquad (\text {C})\ 4010 \qquad (\text {D}) \ 2005^2 \qquad (\text {E})\ 2^{2005}

Solution

For the two functions and f(x)=2-2x,\frac{1}{2}\le x\le 1,we can see that as long as is between and , will be in the right domain. Therefore, we don't need to worry about the domain of . Also, every time we change , the final equation will be in a different form and thus we will get a different value of x. Every time we have two choices for ) and altogether we have to choose times. Thus, .

See Also

2005 AMC 12A (Problems)
Preceded by
Problem 19
Followed by
Problem 21
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Preparing for MATHCOUNTS or the AMC contests, and having a tough time with number theory problems? Read Art of Problem Solving's Introduction to Number Theory by Mathew Crawford.
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