Difference between revisions of "2002 AMC 12P Problems/Problem 2"

(Solution 1)
(Solution 1)
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  n & 0 & 1 & 2 & 3 & 4 & ...\\  
 
  n & 0 & 1 & 2 & 3 & 4 & ...\\  
 
  \hline
 
  \hline
  u_{n} & 4 & 5 & 2 & 1 & 4 & ...\\
+
  uₙ & 4 & 5 & 2 & 1 & 4 & ...\\
 
  \hline
 
  \hline
 
\end{tabular}
 
\end{tabular}

Revision as of 02:34, 30 December 2023

Problem

The function $f$ is given by the table

\[\begin{tabular}{|c||c|c|c|c|c|}  \hline   x & 1 & 2 & 3 & 4 & 5 \\   \hline  f(x) & 4 & 1 & 3 & 5 & 2 \\  \hline \end{tabular}\]

If $u_0=4$ and $u_{n+1} = f(u_n)$ for $n \ge 0$, find $u_{2002}$

$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \text{(E) }5$

Solution 1

We can guess that the series given by the problem is periodic in some way. Starting off, $u_0=4$ is given. $u_1=u_{0+1}=f(u_0)=f(4)=5,$ so $u_1=5.$ $u_1=u_{0+1}=f(u_0)=f(4)=5,$ so $u_1=5.$ $u_2=u_{1+1}=f(u_1)=f(5)=2,$ so $u_2=2.$ $u_3=u_{2+1}=f(u_2)=f(2)=1,$ so $u_3=1.$ $u_4=u_{3+1}=f(u_3)=f(1)=4,$ so $u_4=4.$ Plugging in $4$ will give us $5$ as found before, and plugging in $5$ will give $2$ and so on. This means that our original guess of the series being periodic was correct. Summing up our findings in a nice table,

\[\begin{tabular}{|c||c|c|c|c|c|c|}
 \hline 
 n & 0 & 1 & 2 & 3 & 4 & ...\\ 
 \hline
 uₙ & 4 & 5 & 2 & 1 & 4 & ...\\
 \hline
\end{tabular}\] (Error compiling LaTeX. Unknown error_msg)

in which the next $u_n$ is found by simply plugging in the number from the last box into $f(x).$ The function is periodic every $4 terms$. $2002 \equiv 2\pmod{4}$, and counting $4$ starting from $u_1$ will give us our answer of $\boxed{\textbf{(B) } 2}$.

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
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

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