Difference between revisions of "2015 AMC 12A Problems/Problem 22"
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==Problem== | ==Problem== | ||
− | For each positive integer <math>n</math>, let <math>S(n)</math> be the number of sequences of length <math>n</math> consisting solely of the letters <math>A</math> and <math>B</math>, with no more than three <math>A</math>s in a row and no more than three <math>B</math>s in a row. What is the remainder when <math>S(2015)</math> is divided by 12? | + | For each positive integer <math>n</math>, let <math>S(n)</math> be the number of sequences of length <math>n</math> consisting solely of the letters <math>A</math> and <math>B</math>, with no more than three <math>A</math>s in a row and no more than three <math>B</math>s in a row. What is the remainder when <math>S(2015)</math> is divided by <math>12</math>? |
− | <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D) | + | <math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10 </math> |
− | ==Solution== | + | ==Solution 1== |
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>. | ||
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We can thus begin calculating values of <math>A(n)</math>. We see that the sequence goes (starting from <math>A(0) = 1</math>): <math>1,1,2,4,7,13,24...</math> | We can thus begin calculating values of <math>A(n)</math>. We see that the sequence goes (starting from <math>A(0) = 1</math>): <math>1,1,2,4,7,13,24...</math> | ||
− | A problem arises though | + | A problem arises though: the values of <math>A(n)</math> increase at an exponential rate. Notice however, that we need only find <math>S(2015)\ \text{mod}\ 12</math>. In fact, we can use the fact that <math>S(n) = 2A(n)</math> to only need to find <math>A(2015)\ \text{mod}\ 6</math>. Going one step further, we need only find <math>A(2015)\ \text{mod}\ 2</math> and <math>A(2015)\ \text{mod}\ 3</math> to find <math>A(2015)\ \text{mod}\ 6</math>. |
Here are the values of <math>A(n)\ \text{mod}\ 2</math>, starting with <math>A(0)</math>: <cmath>1,1,0,0,1,1,0,0...</cmath> | Here are the values of <math>A(n)\ \text{mod}\ 2</math>, starting with <math>A(0)</math>: <cmath>1,1,0,0,1,1,0,0...</cmath> | ||
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Knowing that <math>A(2015) \equiv 0\ \text{mod}\ 2</math> and <math>A(2015) \equiv 1\ \text{mod}\ 3</math>, we see that <math>A(2015) \equiv 4\ \text{mod}\ 6</math>, and <math>S(2015) \equiv 8\ \text{mod}\ 12</math>. Hence, the answer is <math>\textbf{(D)}</math>. | Knowing that <math>A(2015) \equiv 0\ \text{mod}\ 2</math> and <math>A(2015) \equiv 1\ \text{mod}\ 3</math>, we see that <math>A(2015) \equiv 4\ \text{mod}\ 6</math>, and <math>S(2015) \equiv 8\ \text{mod}\ 12</math>. Hence, the answer is <math>\textbf{(D)}</math>. | ||
+ | |||
+ | * Note that instead of introducing <math>A(n)</math> and <math>B(n)</math>, we can simply write the relation <math>S(n)=S(n-1)+S(n-2)+S(n-3),</math> and proceed as above. | ||
+ | |||
+ | ==Recursion Solution 2== | ||
+ | The huge <math>n</math> 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 <math>S(n)</math> from previous cases. | ||
+ | So how can we make the words of <math>S(n)</math>? Do we choose 3-in-a-row of one letter, <math>A</math> or <math>B</math>, or do we want <math>2</math> consecutive ones or <math>1</math>? Note that this covers all possible cases of ending with <math>A</math> and <math>B</math> with a certain number of consecutive letters. And obviously they are all distinct. | ||
+ | |||
+ | [Convince yourself that each case for <math>S(n)</math> is considered exactly once by using these cues: does it end in <math>3</math>, <math>2</math>, or <math>1</math> consecutive letter(s) (<math>1</math> consecutive means a string like ...<math>BA</math>, ...<math>AB</math>, as in the letter switches) and does it <math>WLOG</math> consider both <math>A</math> and <math>B?</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)</math> mod 12. We first compute <math>S(2015)</math> mod <math>3</math> and mod <math>4</math>. By listing out the residues mod <math>3</math>, we find that the cycle length for mod <math>3</math> is <math>13</math>. <math>2015</math> is <math>0</math> mod <math>13</math> so <math>S(2015) = S(13) = 2</math> mod <math>3</math>. By listing out the residues mod <math>4</math>, we find that the cycle length for mod <math>4</math> is <math>4</math>. S(2015) = S(3) = mod <math>4</math>. By Chinese Remainder Theorem, <math>S(2015) =\boxed{8}</math> mod <math>12</math>. | ||
+ | |||
+ | ==Solution 3 (Easy Version)== | ||
+ | We can start off by finding patterns in <math>S(n)</math>. 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 <math>S(n) = 2^n - 2((n_4)- (n_5) \dots (n_n))</math>. Rearranging the expression we realize that the terms aside from <math>2^{2015}</math> are congruent to <math>0</math> mod <math>12</math>(Just put the equation in terms of 2^{2015} and the four combinations excluded and calculate the combinations mod <math>12</math>). Using patterns we can see that <math>2^{2015}</math> is congruent to <math>8</math> mod <math>12</math>. Therefore <math>\boxed {8}</math> is our answer. | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2015amc12a/400 | ||
+ | |||
+ | ~ dolphin7 | ||
+ | |||
+ | == See Also == | ||
+ | {{AMC12 box|year=2015|ab=A|num-b=21|num-a=23}} |
Revision as of 21:03, 21 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 |