Difference between revisions of "2015 AMC 12A Problems/Problem 22"
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==Solution 2== | ==Solution 2== | ||
− | We can also go straight to S(2015). We know that there are | + | We can also go straight to S(2015). We know that there are 2^2015 permutations if we do not consider the rule of "no three As or Bs in a row". |
− | Now if the rule is considered, Assume that there the 3 As and 3Bs are in the very front. That would yield | + | Now if the rule is considered, Assume that there the 3 As and 3Bs are in the very front. That would yield (1/2)^3 * 2^2012 * 2 permutations. Now consider where those three As or three Bs can be placed. There are 2013 places where the consecutive letter can be put. Thus there are 2^2015 - 2^2010 * 2013 permutations allowed. |
− | Now do some calculation we can know that <math>2^1 \equiv 2\ \text{mod}\ 12</math>, <math>2^2 \equiv 4\ \text{mod}\ 12</math>, <math>2^3 \equiv 8\ \text{mod}\ 12</math>, ... | + | Now do some calculation we can know that <math>2^1 \equiv 2\ \text{mod}\ 12</math>, <math>2^2 \equiv 4\ \text{mod}\ 12</math>, <math>2^3 \equiv 8\ \text{mod}\ 12</math>, ... 2^2010 \equiv 4\ \text{mod}\ 12, ... 2^2015 \equiv 8\ \text{mod}\ 12. Also, from division, <math>2013 \equiv 9\ \text{mod}\ 12</math>. So <math>2^2015 - 2^2010*2013 \equiv 8\ \text{mod}\ 12</math>. |
Thus, the answer is <math>\textbf{(D)}</math>. | Thus, the answer is <math>\textbf{(D)}</math>. |
Revision as of 11:25, 11 September 2017
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 .
Solution 2
We can also go straight to S(2015). We know that there are 2^2015 permutations if we do not consider the rule of "no three As or Bs in a row".
Now if the rule is considered, Assume that there the 3 As and 3Bs are in the very front. That would yield (1/2)^3 * 2^2012 * 2 permutations. Now consider where those three As or three Bs can be placed. There are 2013 places where the consecutive letter can be put. Thus there are 2^2015 - 2^2010 * 2013 permutations allowed.
Now do some calculation we can know that , , , ... 2^2010 \equiv 4\ \text{mod}\ 12, ... 2^2015 \equiv 8\ \text{mod}\ 12. Also, from division, . So .
Thus, the answer is .
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 |