2001 AMC 12 Problems/Problem 25

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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 $\forall n>1:~ a_n = a_{n-1}a_{n+1} - 1$ . 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:

$\begin{align*}a_3 & = \frac{a_2+1}{a_1} = \frac{2001}x\\a_4 & = \frac{a_3+1}{a_2}= \frac{ \dfrac{2001}x + 1 }{ 2000 }...$

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 $2001$ .

2001 AMC 12 (Problems • 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

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2001 AMC 12 Problems/Problem 25

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