Difference between revisions of "2019 AMC 12B Problems/Problem 23"

m (Solution)
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This is because for any valid sequence of length <math>n</math>, you can remove either the last two 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 two 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 follow the recursion up until <math>f(19) = 65 \quad \boxed{C}</math>
+
Since <math>f(5) = 1</math> and <math>f(6) = 2</math>, you follow the recursion up until <math>f(19) = 65 \quad \boxed{C}</math>
  
 
-Solution by MagentaCobra
 
-Solution by MagentaCobra

Revision as of 11:45, 14 February 2019

Problem

Solution

We can deduce that any valid sequence of length $n$ wil start with a 0 followed by either "10" or "110". Because of this, we can define a recursive function:

$f(n) = f(n-3) + f(n-2)$

This is because for any valid sequence of length $n$, you can remove either the last two 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 follow the recursion up until $f(19) = 65 \quad \boxed{C}$

-Solution by MagentaCobra

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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
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