2015 AMC 12A Problems/Problem 22
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 |