Difference between revisions of "2002 AMC 12P Problems/Problem 2"
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== Solution == | == Solution == | ||
− | 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> | + | 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> |
== See also == | == See also == | ||
{{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}} |
Revision as of 02:06, 30 December 2023
Problem
The function is given by the table
If and
for
, find
Solution
We can guess that the series given by the problem is periodic in some way. Starting off, is given.
so
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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