Difference between revisions of "2019 AMC 10B Problems/Problem 25"
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| + | {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}} | ||
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==Problem== | ==Problem== | ||
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| + | How many sequences of <math>0</math>s and <math>1</math>s of length <math>19</math> are there that begin with a <math>0</math>, end with a <math>0</math>, contain no two consecutive <math>0</math>s, and contain no three consecutive <math>1</math>s? | ||
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| + | <math>\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75</math> | ||
==Solution== | ==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: | ||
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| + | <math>f(n) = f(n-3) + f(n-2)</math> | ||
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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. | ||
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| + | 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> | ||
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| + | -Solution by MagentaCobra | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}} | ||
| + | {{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:27, 14 February 2019
- The following problem is from both the 2019 AMC 10B #25 and 2019 AMC 12B #23, so both problems redirect to this page.
Problem
How many sequences of 
s and 
s of length 
are there that begin with a , end with a , contain no two consecutive s, and contain no three consecutive s?
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 two numbers ("10") or the last three numbers ("110") and the sequence would still satisfy the given conditions.
Since 
and 
, you follow the recursion up until 
-Solution by MagentaCobra
See Also
| 2019 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 10 Problems and Solutions | ||
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
