Difference between revisions of "2002 AMC 12P Problems/Problem 2"
(→Solution 1) |
(→Solution 1) |
||
Line 34: | Line 34: | ||
n & 0 & 1 & 2 & 3 & 4 & ...\\ | n & 0 & 1 & 2 & 3 & 4 & ...\\ | ||
\hline | \hline | ||
− | + | u_{n} & 4 & 5 & 2 & 1 & 4 & ...\\ | |
\hline | \hline | ||
\end{tabular} | \end{tabular} |
Revision as of 02:34, 30 December 2023
Problem
The function is given by the table
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
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,
\[\begin{tabular}{|c||c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & 4 & ...\\ \hline u_{n} & 4 & 5 & 2 & 1 & 4 & ...\\ \hline \end{tabular}\] (Error compiling LaTeX. Unknown error_msg)
in which the next is found by simply plugging in the number from the last box into
The function is periodic every
.
, 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.