2001 AMC 12 Problems/Problem 25

Problem

Consider sequences of positive real numbers of the form $x, 2000, y, \dots$ in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of $x$ does the term 2001 appear somewhere in the sequence?

$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \text{(E) more than }4$

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 $\forall$ (for all) $n>1:~ a_n = a_{n-1}a_{n+1} - 1$. This can be rewritten as $a_{n+1} = \frac{a_n +1}{a_{n-1}}$. We have $a_1=x$ and $a_2=2000$, 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}{2000x} }{ \frac{2001}x } = \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*}

At this point we see that the sequence will become periodic: we have $a_6=a_1$, $a_7=a_2$, and each subsequent term is uniquely determined by the previous two.

Hence if $2001$ appears, it has to be one of $a_1$ to $a_5$. As $a_2=2000$, we only have four possibilities left. Clearly $a_1=2001$ for $x=2001$, and $a_3=2001$ for $x=1$. The equation $a_4=2001$ solves to $x = \frac{2001}{2000\cdot 2001 - 1}$, and the equation $a_5=2001$ to $x=2000\cdot 2001 - 1$.

No two values of $x$ we just computed are equal, and therefore there are $\boxed{4}$ different values of $x$ for which the sequence contains the value $2001$.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
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