2001 AMC 12 Problem 25

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

The condition

b = ac - 1

rearranges to

c = \dfrac{1 + b}{a},

letting you march forward

Doing this repeatedly shows the sequence is periodic with period

5,

so

2001

can only be one of five terms

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. Setting each of the other four displayed terms equal to

2001

gives, respectively,

\begin{gathered} a=2001,\quad a=1,\\ a=\frac{2001}{4001999},\quad a=4001999. \end{gathered}

These four values are positive and distinct. The second term itself is

2000,

and periodicity shows there are no other positions to consider.

So there are

4

values of

x.

Thus, the correct answer is D.

Problem 25 in other years