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2015 AMC 12A Problems/Problem 22

Revision as of 11:25, 11 September 2017 by Hbh567 (talk | contribs) (→‎Solution 2)

Problem

For each positive integer $n$, let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$, with no more than three $A$s in a row and no more than three $B$s in a row. What is the remainder when $S(2015)$ is divided by $12$?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$

Solution

One method of approach is to find a recurrence for $S(n)$.

Let us define $A(n)$ as the number of sequences of length $n$ ending with an $A$, and $B(n)$ as the number of sequences of length $n$ ending in $B$. Note that $A(n) = B(n)$ and $S(n) = A(n) + B(n)$, so $S(n) = 2A(n)$.

For a sequence of length $n$ ending in $A$, it must be a string of $A$s appended onto a sequence ending in $B$ of length $n-1, n-2, \text{or}\ n-3$. So we have the recurrence: \[A(n) = B(n-1) + B(n-2) + B(n-3) = A(n-1) + A(n-2) + A(n-3)\]

We can thus begin calculating values of $A(n)$. We see that the sequence goes (starting from $A(0) = 1$): $1,1,2,4,7,13,24...$

A problem arises though: the values of $A(n)$ increase at an exponential rate. Notice however, that we need only find $S(2015)\ \text{mod}\ 12$. In fact, we can use the fact that $S(n) = 2A(n)$ to only need to find $A(2015)\ \text{mod}\ 6$. Going one step further, we need only find $A(2015)\ \text{mod}\ 2$ and $A(2015)\ \text{mod}\ 3$ to find $A(2015)\ \text{mod}\ 6$.

Here are the values of $A(n)\ \text{mod}\ 2$, starting with $A(0)$: \[1,1,0,0,1,1,0,0...\]

Since the period is $4$ and $2015 \equiv 3\ \text{mod}\ 4$, $A(2015) \equiv 0\ \text{mod}\ 2$.

Similarly, here are the values of $A(n)\ \text{mod}\ 3$, starting with $A(0)$: \[1,1,2,1,1,1,0,2,0,2,1,0,0,1,1,2...\]

Since the period is $13$ and $2015 \equiv 0\ \text{mod}\ 13$, $A(2015) \equiv 1\ \text{mod}\ 3$.

Knowing that $A(2015) \equiv 0\ \text{mod}\ 2$ and $A(2015) \equiv 1\ \text{mod}\ 3$, we see that $A(2015) \equiv 4\ \text{mod}\ 6$, and $S(2015) \equiv 8\ \text{mod}\ 12$. Hence, the answer is $\textbf{(D)}$.


Solution 2

We can also go straight to S(2015). We know that there are 2^2015 permutations if we do not consider the rule of "no three As or Bs in a row".

Now if the rule is considered, Assume that there the 3 As and 3Bs are in the very front. That would yield (1/2)^3 * 2^2012 * 2 permutations. Now consider where those three As or three Bs can be placed. There are 2013 places where the consecutive letter can be put. Thus there are 2^2015 - 2^2010 * 2013 permutations allowed.

Now do some calculation we can know that $2^1 \equiv 2\ \text{mod}\ 12$, $2^2 \equiv 4\ \text{mod}\ 12$, $2^3 \equiv 8\ \text{mod}\ 12$, ... 2^2010 \equiv 4\ \text{mod}\ 12, ... 2^2015 \equiv 8\ \text{mod}\ 12. Also, from division, $2013 \equiv 9\ \text{mod}\ 12$. So $2^2015 - 2^2010*2013 \equiv 8\ \text{mod}\ 12$.

Thus, the answer is $\textbf{(D)}$.

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
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