Difference between revisions of "2015 AMC 12A Problems/Problem 22"
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Let us define <math>A(n)</math> as the number of sequences of length <math>n</math> ending with an <math>A</math>, and <math>B(n)</math> as the number of sequences of length <math>n</math> ending in <math>B</math>. Note that <math>A(n) = B(n)</math> and <math>S(n) = A(n) + B(n)</math>, so <math>S(n) = 2A(n)</math>. | Let us define <math>A(n)</math> as the number of sequences of length <math>n</math> ending with an <math>A</math>, and <math>B(n)</math> as the number of sequences of length <math>n</math> ending in <math>B</math>. Note that <math>A(n) = B(n)</math> and <math>S(n) = A(n) + B(n)</math>, so <math>S(n) = 2A(n)</math>. |
Revision as of 00:28, 5 February 2015
Problem
For each positive integer , let
be the number of sequences of length
consisting solely of the letters
and
, with no more than three
s in a row and no more than three
s in a row. What is the remainder when
is divided by 12?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}}\ 8\qquad\textbf{(E)}\ 10$ (Error compiling LaTeX. Unknown error_msg)
Solution
One method of approach is to find a recurrence for .
Let us define as the number of sequences of length
ending with an
, and
as the number of sequences of length
ending in
. Note that
and
, so
.
For a sequence of length ending in
, it must be a string of
s appended onto a sequence of length
. So we have the recurrence
We can thus begin calculating values of . We see that the sequence goes (starting from
):
A problem arises though - the values soon get to be far to large. Notice however, that we need only find . In fact, we can abuse the fact that
and only find
. Going one step further, we need only find the period of
and
to find
.
Here are the values of , starting with
:
Since the period is and
,
.
Similarly, here are the values of , starting with
:
Since the period is and
,
.
Knowing that and
, we see that
, and
. Hence, the answer is
.