Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 12P Problems/Problem 2"
(→‎See also)
 
(24 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #2]] and [[2002 AMC 10P Problems|2002 AMC 10P #9]]}}
 +
 
==Problem==
 
==Problem==
 
The function <math>f</math> is given by the table
 
The function <math>f</math> is given by the table
Line 26: Line 28:
 
</math>
 
</math>
  
−
== Solution ==
+
== Solution 1==
−
If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
+
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>
 +
\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}
 +
</cmath>
 +
 
 +
in which the next <math>u_n</math> is found by simply plugging in the number from the last box into <math>f(x).</math> The function is periodic every <math>4</math> terms. <math>2002 \equiv 2\pmod{4}</math>, and counting <math>4</math> starting from <math>u_1</math> will give us our answer of <math>\boxed{\textbf{(B) } 2}</math>.
  
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2002|ab=P|num-b=8|num-a=10}}
 
{{AMC12 box|year=2002|ab=P|num-b=1|num-a=3}}
 
{{AMC12 box|year=2002|ab=P|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:20, 14 July 2024

The following problem is from both the 2002 AMC 12P #2 and 2002 AMC 10P #9, so both problems redirect to this page.

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 10P (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 10 Problems and Solutions
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_2&oldid=222475"