Difference between revisions of "2019 AMC 12B Problems/Problem 23"
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This is because for any valid sequence of length <math>n</math>, you can remove either the last 2 numbers ("10) or the last three numbers ("110") and the sequence would still satisfy the given conditions. | This is because for any valid sequence of length <math>n</math>, you can remove either the last 2 numbers ("10) or the last three numbers ("110") and the sequence would still satisfy the given conditions. | ||
| − | Since f(5) = 1 and f(6) = 2, you build up until <math>f(19) = 65 \boxed{C}</math> | + | Since f(5) = 1 and f(6) = 2, you build up until <math>f(19) = 65 \tab \boxed{C}</math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}} | ||
Revision as of 12:36, 14 February 2019
Problem
Solution
We can deduce that any valid sequence of length 
wil start with a 0 followed by either "10" or "110".
Because of this, we can define a recursive function:
This is because for any valid sequence of length , you can remove either the last 2 numbers ("10) or the last three numbers ("110") and the sequence would still satisfy the given conditions.
Since f(5) = 1 and f(6) = 2, you build up until $f(19) = 65 \tab \boxed{C}$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 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 | |
