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. We need to find <math>S(2015) | + | 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 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 <math>S(2015) = S(13) = 2</math> 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})<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 , 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. We need to find mod 12. We first compute 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 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})=\boxed{8}.
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.
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 |