2001 AMC 12 Problems/Problem 25
Problem
Consider sequences of positive real numbers of the form in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of does the term 2001 appear somewhere in the sequence?
Solution
It never hurts to compute a few terms of the sequence in order to get a feel how it looks like. In our case, the definition is that (for all) . This can be rewritten as . We have and , and we compute:
At this point we see that the sequence will become periodic: we have , , and each subsequent term is uniquely determined by the previous two.
Hence if appears, it has to be one of to . As , we only have four possibilities left. Clearly for , and for . The equation solves to , and the equation to .
No two values of we just computed are equal, and therefore there are different values of for which the sequence contains the value .
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
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