Difference between revisions of "2015 AMC 12A Problems/Problem 22"
(→Solution) |
(→Solution) |
||
Line 27: | Line 27: | ||
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. | |
− | * Note that instead of | ||
== 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 22:57, 3 December 2017
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.
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 |