2009 AMC 12A Problems/Problem 25
Problem
The first two terms of a sequence are and . For ,
What is ?
Solution 1
Consider another sequence such that , and .
The given recurrence becomes
It follows that . Since , all terms in the sequence will be a multiple of .
Now consider another sequence such that , and . The sequence satisfies .
As the number of possible consecutive two terms is finite, we know that the sequence is periodic. Write out the first few terms of the sequence until it starts to repeat.
Note that and . Thus has a period of : .
It follows that and . Thus
Our answer is .
Solution 2
First, some intuition. The given recurrence relation: looks much like the tangent addition formula. So, we let . Then, we get: This gives us: Now, observe that We know that this sequence of values for will repeat eventually because it is mod . It is just a matter of when, so we start bashing:
And there is the repetition. So, this series has a period of 24. , so ~Firebolt360
Clarification
(While the above solution is quite intuitive, it's kinda hard for people who don't know many proofing symbols to understand.)
Notice that the formula given in the problem looks incredibly similar to the Tangent Addition Formula.
Since , let be . Similarly, let be . Then the formula for reads
But from the Tangent Addition Formula we know that this is just the formula for or , meaning . So, the sequence becomes a sort of Fibonacci sequence with angle measures. We continue to sum angle measures, like so:
... wait a minute! !
So now we have
.... Ok this is getting tiring. Let's stop and think about whether we can simplify this. Notice that all of the angle measures are a multiple of . So, let's express the angle measures as multiples of .
Let .
(Basically, is the angle measure of the corresponding divided by )
Now we have
But wait... we're dealing with the function, which has a period (recurrence rate) of or . Since we divided the angle measures by , the period is now (which aligns with what we got earlier: ). This means that we can reduce the terms of the sequence based on the function, which returns the remainder after dividing by a certain amount. So now we can say ( denotes modular equivalence in this context). Now we continue our sequence:
(It's starting to repeat now)
So , and so on. The sequence repeats every terms. The problem asks us for the value of . Let's whip out the function again.
So we want the value of the th number in the sequence, and looking back at it (hopefully you wrote it down somewhere) we find that .
See also
2009 AMC 12A (Problems • Answer Key • Resources) | |
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