Difference between revisions of "2002 AMC 12P Problems/Problem 2"
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+ | {{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #2]] and [[2002 AMC 10P Problems|2002 AMC 10P #9]]}} | ||
+ | |||
+ | ==Problem== | ||
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 1== | ||
+ | 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 21:20, 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 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 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,
in which the next is found by simply plugging in the number from the last box into The function is periodic every terms. , and counting starting from will give us our answer of .
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.