Difference between revisions of "2019 AMC 12B Problems/Problem 23"
Magentacobra (talk | contribs) (→Solution) |
Magentacobra (talk | contribs) (→Solution) |
||
Line 9: | Line 9: | ||
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 \ | + | Since f(5) = 1 and f(6) = 2, you build up until <math>f(19) = 65 \quad \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 12:37, 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
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 |