Difference between revisions of "2002 AMC 12P Problems/Problem 2"
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The function <math>f</math> is given by the table | The function <math>f</math> is given by the table | ||
| − | < | + | <cmath> |
\begin{tabular}{|c||c|c|c|c|c|} | \begin{tabular}{|c||c|c|c|c|c|} | ||
\hline | \hline | ||
x & 1 & 2 & 3 & 4 & 5 \\ | x & 1 & 2 & 3 & 4 & 5 \\ | ||
\hline | \hline | ||
| − | + | f(x) & 4 & 1 & 3 & 5 & 2 \\ | |
\hline | \hline | ||
\end{tabular} | \end{tabular} | ||
| − | + | </cmath> | |
If <math>u_0=4</math> and <math>u_{n+1} = f(u_n)</math> for <math>n \ge 0</math>, find <math>u_{2002}</math> | If <math>u_0=4</math> and <math>u_{n+1} = f(u_n)</math> for <math>n \ge 0</math>, find <math>u_{2002}</math> | ||
Revision as of 17:04, 1 July 2024
Problem
The function 
is given by the table
![\[\begin{tabular}{|c||c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 4 & 1 & 3 & 5 & 2 \\ \hline \end{tabular}\]](http://latex.artofproblemsolving.com/8/3/c/83c499dfa0e4f75b195d53d345895ba9dec1cc50.png)
If 
and 
for 
, find 
Solution 1
We can guess that the series given by the problem is periodic in some way. Starting off, is given. 
so 
so 
so 
so 
Plugging in 
will give us 
as found before, and plugging in will give 
and so on. This means that our original guess of the series being periodic was correct. Summing up our findings in a nice table,
![\[\begin{tabular}{|c||c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & 4 & ...\\ \hline un & 4 & 5 & 2 & 1 & 4 & ...\\ \hline \end{tabular}\]](http://latex.artofproblemsolving.com/5/e/9/5e928916666d2dda563e44b0cca4adc1fb51d27e.png)
in which the next 
is found by simply plugging in the number from the last box into 
The function is periodic every terms. 
, and counting starting from 
will give us our answer of 
.
See also
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
