Difference between revisions of "2001 AMC 12 Problems/Problem 25"
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Revision as of 20:05, 3 July 2013
Problem
Consider sequences of positive real numbers of the form 
in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of 
does the term 2001 appear somewhere in the sequence?
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 
. This can be rewritten as 
. We have 
and 
, 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} }{ 2000\cdot 2001 }
= \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*} (Error compiling LaTeX. Unknown error_msg)
At this point we see that the sequence will become periodic: we have 
, 
, and each subsequent term is uniquely determined by the previous two.
Hence if 
appears, it has to be one of 
to 
. As , we only have four possibilities left. Clearly 
for 
, and 
for 
. The equation 
solves to 
, and the equation 
to 
.
No two values of we just computed are equal, and therefore there are 
different values of for which the sequence contains the value .
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
