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

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 II

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. Let's start using all $S(n)$ values $(\mod12)$ . We also know that $S(1)=2, S(2)=4, S(3)=8$ , and so on. These residues are: $2, 4, 8, 2, 2, 0, 4, 6, 10, 8, 0, 8, 8, 4, 8, 8, 8, 0...$ , upon which the cycle repeats. Note that the cycle length is 7, and $2015-286*7=2002=13$ , so the residue of $S(2015) (\mod{12})$ is the residue of $S(13)=\boxed{008}$ .

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2015 AMC 12A Problems/Problem 22

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Problem

Solution

Recursion Solution II

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