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 . This can be rewritten as
. We have
and
, and we compute:
\begin{align*} a_3 & = \frac{a_2+1}{a_1} = \frac{2001}x \\ a_4 & = \frac{a_3+1}{a_2} = \frac{ \dfrac{2001}x + 1 }{ 2000 } = \frac{2001 + x}{2000x} \\ a_5 & = \frac{a_4+1}{a_3} = \frac{ \frac{2001 + x}{2000x} + 1 }{ \frac{2001}x } = \frac{ \frac{2001 + 2001x} }{ 2000\cdot 2001 } = \frac{1+x}{2000} \\ a_6 & = \frac{a_5+1}{a_4} = \frac{ \frac{1+x}{2000} + 1 }{ \frac{2001 + x}{2000x} } = \frac{ \frac{2001+x}{2000} }{ \frac{2001 + x}{2000x} } = x \\ a_7 & = \frac{a_6+1}{a_5} = \frac{ x+1 }{ \frac{1+x}{2000} } = 2000 \end{align*} (Error compiling LaTeX. Unknown error_msg)
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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