Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2001 AMC 12 Problems/Problem 25

Revision as of 02:09, 5 November 2024 by Tsun26 (talk | contribs) (→‎Remark)

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 $2001$.

Remark

In general, notice that the sequence defined by $a_{n+1} = \frac{a_n +1}{a_{n-1}}$ for $n\ge 2$ and $a_1=a, a_2=b$ is periodic.

~tsun26

See Also

2001 AMC 12 (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_25&oldid=231655"