2015 AMC 12A Problems/Problem 22

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 1

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 $\textbf{(D)}$ .

Note that instead of introducing $A(n)$ and $B(n)$ , we can simply write the relation $S(n)=S(n-1)+S(n-2)+S(n-3),$ and proceed as above.

Recursion Solution 2

The huge $n$ 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 $S(n)$ from previous cases. So how can we make the words of $S(n)$ ? Do we choose 3-in-a-row of one letter, $A$ or $B$ , or do we want $2$ consecutive ones or $1$ ? Note that this covers all possible cases of ending with $A$ and $B$ with a certain number of consecutive letters. And obviously they are all distinct.

[Convince yourself that each case for $S(n)$ is considered exactly once by using these cues: does it end in $3$ , $2$ , or $1$ consecutive letter(s) ( $1$ consecutive means a string like ... $BA$ , ... $AB$ , as in the letter switches) and does it $WLOG$ consider both $A$ and $B?$ ]

From there we realize that $S(n)=S(n-1)+S(n-2)+S(n-3)$ because 3 in a row requires $S(n-3)$ , and so on. We need to find $S(2015)$ mod 12. We first compute $S(2015)$ mod $3$ and mod $4$ . By listing out the residues mod $3$ , we find that the cycle length for mod $3$ is $13$ . $2015$ is $0$ mod $13$ so $S(2015) = S(13) = 2$ mod $3$ . By listing out the residues mod $4$ , we find that the cycle length for mod $4$ is $4$ . S(2015) = S(3) = mod $4$ . By Chinese Remainder Theorem, $S(2015) =\boxed{8}$ mod $12$ .

Solution 3 (Easy Version)

We can start off by finding patterns in $S(n)$ . 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 $S(n) = 2^n - 2((n_4)- (n_5) \dots (n_n))$ . Rearranging the expression we realize that the terms aside from $2^{2015}$ are congruent to $0$ mod $12$ (Just put the equation in terms of 2^{2015} and the four combinations excluded and calculate the combinations mod $12$ ). Using patterns we can see that $2^{2015}$ is congruent to $8$ mod $12$ . Therefore $\boxed {8}$ is our answer.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2015amc12a/400

~ dolphin7

2015 AMC 12A (Problems • Answer Key • Resources)
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