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 ( | + | ==Solution 2 (A Bit Bashy)== |
Start out by listing some terms of the sequence. | Start out by listing some terms of the sequence. | ||
<cmath>f(1)=1</cmath> | <cmath>f(1)=1</cmath> | ||
Line 48: | Line 48: | ||
<cmath>f(2018)=\boxed{2017}.</cmath> | <cmath>f(2018)=\boxed{2017}.</cmath> | ||
− | Written | + | Written By: pi3141592 |
==See Also== | ==See Also== |
Revision as of 01:32, 18 February 2018
Problem
A function is defined recursively by and for all integers . What is ?
Solution 1
Thus, .
Solution 2 (A Bit Bashy)
Start out by listing some terms of the sequence.
Notice that whenever is an odd multiple of , and the pattern of numbers that follow will always be +3, +2, +0, -1, +0. The closest odd multiple of to is , so we have
Written By: pi3141592
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 |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.