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Difference between revisions of "2015 AMC 12A Problems/Problem 22"

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[Convince yourself that each case for <math>S(n)</math> 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?]
 
[Convince yourself that each case for <math>S(n)</math> 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?]
  
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From there we realize that <math>S(n)=S(n-1)+S(n-2)+S(n-3)</math> because 3 in a row requires <math>S(n-3)</math>, and so on. We need to find <math>S(2015) (\mod{12})</math>. We first compute <math>S(2015) (\mod {3})</math> and <math>(\mod {4})</math>. By listing out the residues <math>(\mod {3})</math>, we find that the cycle length for <math>(\mod {3})</math> is 13. 2015 is 0 (\mod {13})<math> so </math>S(2015) = S(13) = 2 (\mod {3})<math></math>. By listing out the residues (\mod {4})<math>, we find that the cycle length for (\mod {4})</math> is <math>4</math>. S(2015) = S(3) = (\mod {4})<math>. By Chinese Remainder Theorem, S(2015) </math>=\boxed{8}<math> (\mod {12})</math>.
+
From there we realize that <math>S(n)=S(n-1)+S(n-2)+S(n-3)</math> because 3 in a row requires <math>S(n-3)</math>, and so on. We need to find <math>S(2015) mod 12</math>. We first compute <math>S(2015) mod 3</math> and mod 4. By listing out the residues <math>(\mod {3})</math>, we find that the cycle length for <math>(\mod {3})</math> is 13. 2015 is 0 (\mod {13})<math> so </math>S(2015) = S(13) = 2 (\mod {3})<math></math>. By listing out the residues (\mod {4})<math>, we find that the cycle length for (\mod {4})</math> is <math>4</math>. S(2015) = S(3) = (\mod {4})<math>. By Chinese Remainder Theorem, S(2015) </math>=\boxed{8}<math> (\mod {12})</math>.
  
 
==Solution 3 (Easy Version)==
 
==Solution 3 (Easy Version)==

Revision as of 14:41, 26 January 2020

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. 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})$$ (Error compiling LaTeX. Unknown error_msg). 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.

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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