Difference between revisions of "2018 AMC 10B Problems/Problem 20"
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Thus, <math>f\left(2018\right) = 2016 + f\left(2\right) = 2017</math>. <math>\boxed{B}</math> | Thus, <math>f\left(2018\right) = 2016 + f\left(2\right) = 2017</math>. <math>\boxed{B}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let us write down the first couple terms of the sequence: | ||
+ | <math>f(1)=1</math> | ||
+ | <math>f(2)=1</math> | ||
+ | $f(3) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:11, 16 February 2018
Contents
Problem
A function is defined recursively by and for all integers . What is ?
Solution
Thus, .
Solution 2
Let us write down the first couple terms of the sequence: $f(3)
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
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