Difference between revisions of "2015 AMC 12A Problems/Problem 22"
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For each positive integer <math>n</math>, let <math>S(n)</math> be the number of sequences of length <math>n</math> consisting solely of the letters <math>A</math> and <math>B</math>, with no more than three <math>A</math>s in a row and no more than three <math>B</math>s in a row. What is the remainder when <math>S(2015)</math> is divided by <math>12</math>? | For each positive integer <math>n</math>, let <math>S(n)</math> be the number of sequences of length <math>n</math> consisting solely of the letters <math>A</math> and <math>B</math>, with no more than three <math>A</math>s in a row and no more than three <math>B</math>s in a row. What is the remainder when <math>S(2015)</math> is divided by <math>12</math>? | ||
− | <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D) | + | <math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10 </math> |
==Solution== | ==Solution== |
Revision as of 11:35, 10 March 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
?
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
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
.
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 |