Difference between revisions of "2005 AMC 12A Problems/Problem 20"

(See also)
Line 8: Line 8:
 
* [[2005 AMC 12A Problems/Problem 21 | Next problem]]
 
* [[2005 AMC 12A Problems/Problem 21 | Next problem]]
 
* [[2005 AMC 12A Problems]]
 
* [[2005 AMC 12A Problems]]
 +
For the two functions f(x)=2x (0<=x<=1/2) and f(x)=2-2x (1/2<x<=1),we can see that as long as f(x) is between 0 and 1, x will be in the right domain.  Therefore, we don't need to worry about the domain of x.  Also, every time we change f(x), 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 f(x) and altogether we have 2005 choices. 
 +
2*2*2...2=2^2005

Revision as of 23:28, 24 December 2007

This is an empty template page which needs to be filled. You can help us out by finding the needed content and editing it in. Thanks.

Problem

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

For the two functions f(x)=2x (0<=x<=1/2) and f(x)=2-2x (1/2<x<=1),we can see that as long as f(x) is between 0 and 1, x will be in the right domain. Therefore, we don't need to worry about the domain of x. Also, every time we change f(x), 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 f(x) and altogether we have 2005 choices. 2*2*2...2=2^2005