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

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

Revision as of 20:57, 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 $u_1=5.$ $u_2=u_{1+1}=f(u_1)=f(5)=2,$ 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,

\[\begin{tabular}{|c||c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & 4 & ...\\ \hline un & 4 & 5 & 2 & 1 & 4 & ...\\ \hline \end{tabular}\]

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 (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

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