Difference between revisions of "1984 AIME Problems/Problem 7"
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<math></math>\begin{align*}f(999)=f(f(1004))=f(1001)=998\ | <math></math>\begin{align*}f(999)=f(f(1004))=f(1001)=998\ | ||
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f(998)=f(f(1003))=f(1000)=997 | f(998)=f(f(1003))=f(1000)=997 | ||
f(997)=f(f(1002))=f(999)=998 | f(997)=f(f(1002))=f(999)=998 | ||
Revision as of 18:28, 25 July 2011
Contents
[hide]Problem
The function f is defined on the set of integers and satisfies 
Find 
.
Solution 1
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:
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 
:
![\[f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}\]](http://latex.artofproblemsolving.com/f/b/8/fb80f9ca09d81d49b8754df2ae578bf487bb2f1b.png)
Solution 2
We start by finding values of the function right under 1000 since they require little repetition.
$$ (Error compiling LaTeX. Unknown error_msg)\begin{align*}f(999)=f(f(1004))=f(1001)=998\
f(998)=f(f(1003))=f(1000)=997 f(997)=f(f(1002))=f(999)=998 f(996)=f(f(1001))=f(998)=997\
Soon we realize the 
for integers 
either equal 
or 
based on it parity. (If short on time, a guess of 998 or 997 can be taken now.)
If 
is even 
if is odd 
. 
has even parity, so 
.
See also
| 1984 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
