Difference between revisions of "2001 AMC 12 Problems/Problem 25"
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& = \frac{a_4+1}{a_3} | & = \frac{a_4+1}{a_3} | ||
= \frac{ \frac{2001 + x}{2000x} + 1 }{ \frac{2001}x } | = \frac{ \frac{2001 + x}{2000x} + 1 }{ \frac{2001}x } | ||
− | = \frac{ \frac{2001 + 2001x}{ | + | = \frac{ \frac{2001 + 2001x}{2000x} }{ 2000\cdot 2001 } |
= \frac{1+x}{2000} | = \frac{1+x}{2000} | ||
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Revision as of 11:46, 19 February 2015
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 . 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) | |
Preceded by Problem 24 |
Followed by Last Question |
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 |
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