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|} | ||
| + | \hline | ||
| + | x & 1 & 2 & 3 & 4 & 5 \\ | ||
| + | \hline | ||
| + | f(x) & 4 & 1 & 3 & 5 & 2 \\ | ||
| + | \hline | ||
| + | \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> | ||
| − | <math>\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \text{(E) }5</math> | + | <math> |
| + | \text{(A) }1 | ||
| + | \qquad | ||
| + | \text{(B) }2 | ||
| + | \qquad | ||
| + | \text{(C) }3 | ||
| + | \qquad | ||
| + | \text{(D) }4 | ||
| + | \qquad | ||
| + | \text{(E) }5 | ||
| + | </math> | ||
== Solution == | == Solution == | ||
Revision as of 00:25, 30 December 2023
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/8/3/4/8341eda6fac42d7213b9c03de00ae11811db6a2f.png)
If 
and 
for 
, find 
Solution
If 
, then 
. Since 
, 
must be 
to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values 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. 
