2005 AMC 12A Problems/Problem 20
Contents
Problem
For each in , define Let , and for each integer . For how many values of in is ?
Solution 1
For the two functions and ,as long as is between and , will be in the right domain, so we don't need to worry about the domain of .
Also, every time we change , the expression for the final answer in terms of will be in a different form(although they'll all satisfy the final equation), so we get a different starting value of . Every time we have two choices for ) and altogether we have to choose times. Thus, .
Note: the values of x that satisfy are , , , ,.
Solution 2
We are given that . Thus, . Let be equal to . Thus or or . Now we know is equal to or . Now we know that or . Now we solve for and let . Thus is equal to ,,,and . As we see, has 1 solution, has 2 solutions, and has 4 solutions. Thus for each iteration we double the number of possible solutions. There are 2005 iterations and thus the number of solutions is (not rigorous)
See Also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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