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?] | ||
− | 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. Let's start using all <math>S(n)</math> values <math>(\mod12)</math>. We also know that <math>S(1)=2, S(2)=4, S(3)=8</math>, and so on. These residues are: <math>2, 4, 8, 2, 2, 0, 4, 6, 10, 8, 0, 8, 8, 4, 8, 8, 8, 0...</math>, upon which the cycle repeats. Note that the cycle length is 7, and <math>2015-286*7=2002=13</math>, so the residue of <math>S(2015) (\mod{12})</math> is the residue of <math>S(13)=\boxed{008}</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. Let's start using all <math>S(n)</math> values <math>(\mod12)</math>. We also know that <math>S(1)=2, S(2)=4, S(3)=8</math>, and so on. These residues are: <math>2, 4, 8, 2, 2, 0, 4, 6, 10, 8, 0, 8, 8, 4, 8, 8, 8, 0...</math>, upon which the cycle repeats. Note that the cycle length is 7, and <math>2015-(286*7=2002)=13</math>, so the residue of <math>S(2015) (\mod{12})</math> is the residue of <math>S(13)=\boxed{008}</math>. |
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=21|num-a=23}} | {{AMC12 box|year=2015|ab=A|num-b=21|num-a=23}} |
Revision as of 19:56, 14 February 2018
Contents
[hide]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
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 II
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 2 consecutive ones or 1? 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 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 because 3 in a row requires , and so on. Let's start using all values . We also know that , and so on. These residues are: , upon which the cycle repeats. Note that the cycle length is 7, and , so the residue of is the residue of .
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 |