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

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m (Solution)
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<math>f(n) = f(n-3) + f(n-2)</math>
 
<math>f(n) = f(n-3) + f(n-2)</math>
  
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 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 f(5) = 1 and f(6) = 2, you follow the recursion up until <math>f(19) = 65 \quad \boxed{C}</math>

Revision as of 11:44, 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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