Difference between revisions of "2007 AMC 12A Problems/Problem 11"
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 02:31, 11 September 2007
Problem
The function f is defined on the set of integers and satisfies
Find .
Solution
Define , where the function
is performed
times. We find that
.
. So we now need to reduce
.
Let’s write out a couple more iterations of this function:
![$\displaystyle f^{185}(1004) = f^{184}(1001) = f^{183}(998)$](http://latex.artofproblemsolving.com/f/0/2/f021a55297891425b24b2d413ab30750ded0c659.png)
![$= f^{184}(1003) = f^{183}(1000) = f^{182}(997)$](http://latex.artofproblemsolving.com/7/a/8/7a80f348127aa35c61a3edbd57cb8fe21925e967.png)
![$= f^{183}(1002) = f^{182}(999) = f^{183}(1004)$](http://latex.artofproblemsolving.com/7/1/1/711b2c1e09a9d998f08affbcd6cfb167592d44cb.png)
So this function reiterates with a period of 2 for . It might be tempting at first to assume that
is the answer; however, that is not true since the solution occurs slightly before that. Start at
:
![$f^{3}(1004) = f^{2}(1001) = f(998)$](http://latex.artofproblemsolving.com/c/3/1/c31b2c1b74b0990abf00cc0fd322cb3cdf890656.png)
![$= f^{2}(1003) = f(1000) = 997$](http://latex.artofproblemsolving.com/0/e/0/0e02a85b0dec5f09d8fa3659cdb6a6b4c6cbe496.png)
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 |