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

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(Redirected page to 2019 AMC 10B Problems/Problem 25)
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==Problem==
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#REDIRECT[[2019_AMC_10B_Problems/Problem_25]]
 
 
==Solution==
 
We can deduce that any valid sequence of length <math>n</math> wil start with a 0 followed by either "10" or "110".
 
Because of this, we can define a recursive function:
 
 
 
<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 two numbers ("10") or the last three numbers ("110") and the sequence would still satisfy the given conditions.
 
 
 
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
 
 
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}
 

Latest revision as of 13:27, 14 February 2019