Difference between revisions of "2015 AMC 12A Problems/Problem 22"
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<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10 </math> | <math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10 </math> | ||
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One method of approach is to find a recurrence for <math>S(n)</math>. | One method of approach is to find a recurrence for <math>S(n)</math>. |
Revision as of 16:21, 20 June 2020
Contents
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 1
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 use the fact that
to only need to 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
.
- Note that instead of introducing
and
, we can simply write the relation
and proceed as above.
Recursion Solution 2
The huge value in place, as well as the "no more than... in a row" are key phrases that indicate recursion is the right way to go.
Let's go with finding the case of
from previous cases.
So how can we make the words of
? Do we choose 3-in-a-row of one letter,
or
, or do we want
consecutive ones or
? Note that this covers all possible cases of ending with
and
with a certain number of consecutive letters. And obviously they are all distinct.
[Convince yourself that each case for is considered exactly once by using these cues: does it end in
,
, or
consecutive letter(s) (
consecutive means a string like ...
, ...
, as in the letter switches) and does it
consider both
and
]
From there we realize that because 3 in a row requires
, and so on. We need to find
mod 12. We first compute
mod
and mod
. By listing out the residues mod
, we find that the cycle length for mod
is
.
is
mod
so
mod
. By listing out the residues mod
, we find that the cycle length for mod
is
. S(2015) = S(3) = mod
. By Chinese Remainder Theorem,
mod
.
Solution 3 (Easy Version)
We can start off by finding patterns in . When we calculate a few values we realize either from performing the calculation or because the calculation was performed in the exact same way that
. Rearranging the expression we realize that the terms aside from
are congruent to
mod
(Just put the equation in terms of 2^{2015} and the four combinations excluded and calculate the combinations mod
). Using patterns we can see that
is congruent to
mod
. Therefore
is our answer.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc12a/400
~ dolphin7
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 |