Difference between revisions of "2015 AMC 12A Problems/Problem 22"
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We can thus begin calculating values of <math>A(n)</math>. We see that the sequence goes (starting from <math>A(0) = 1</math>): <math>1,1,2,4,7,13,24...</math> | We can thus begin calculating values of <math>A(n)</math>. We see that the sequence goes (starting from <math>A(0) = 1</math>): <math>1,1,2,4,7,13,24...</math> | ||
− | A problem arises though | + | A problem arises though: the values of <math>A(n)</math> increase at an exponential rate. Notice however, that we need only find <math>S(2015)\ \text{mod}\ 12</math>. In fact, we can abuse the fact that <math>S(n) = 2A(n)</math> and only find <math>A(2015)\ \text{mod}\ 6</math>. Going one step further, we need only find the period of <math>A(2015)\ \text{mod}\ 2</math> and <math>A(2015)\ \text{mod}\ 3</math> to find <math>A(2015)\ \text{mod}\ 6</math>. |
Here are the values of <math>A(n)\ \text{mod}\ 2</math>, starting with <math>A(0)</math>: <cmath>1,1,0,0,1,1,0,0...</cmath> | Here are the values of <math>A(n)\ \text{mod}\ 2</math>, starting with <math>A(0)</math>: <cmath>1,1,0,0,1,1,0,0...</cmath> |
Revision as of 00:31, 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 ending in
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 of increase at an exponential rate. 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
.