Question

Consider sequences of positive real numbers of the form $x, 2000, y, \ldots,$ 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?

$1$

$2$

$3$

$4$

more than $4$

Solution:

If $a, b, c$ are consecutive terms then $b = ac - 1,$ so $c = \dfrac{1 + b}{a}.$ Applying this repeatedly, the first five terms are $a,\ b,\ \dfrac{1 + b}{a},\ \dfrac{1 + a + b}{ab},\ \dfrac{1 + a}{b},$ after which $a$ and $b$ recur, so the sequence is periodic with period $5.$

Here $b = 2000$ is the second term. The value $2001$ can be placed in any one of the other four of the five distinct positions, and each choice determines $x = a$ uniquely and yields a valid sequence of positive reals.

So there are $4$ values of $x.$

Thus, the correct answer is D.

2001 AMC 12 Problem 25

Solution:

Problem 25 in Other Years