Difference between revisions of "2001 AMC 12 Problems/Problem 25"
(New page: == Problem == Consider sequences of positive real numbers of the form <math>x, 2000, y, \dots</math> in which every term after the first is 1 less than the product of its two immediate ne...) |
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== Problem == | == Problem == | ||
− | Consider sequences of positive real numbers of the form <math>x, 2000, y, \dots</math> in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of <math>x</math> does the term 2001 appear somewhere in the sequence? | + | Consider sequences of positive real numbers of the form <math>x, 2000, y, \dots</math> in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of <math>x</math> does the term <math>2001</math> appear somewhere in the sequence? |
<math> | <math> | ||
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== Solution == | == 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 <math>\forall n>1:~ a_n = a_{n-1}a_{n+1} - 1</math>. This can be rewritten as <math>a_{n+1} = \frac{a_n +1}{a_{n-1}}</math>. We have <math>a_1=x</math> and <math>a_2=2000</math>, and we compute: | + | 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 <math>\forall</math> (for all) <math> n>1:~ a_n = a_{n-1}a_{n+1} - 1</math>. This can be rewritten as <math>a_{n+1} = \frac{a_n +1}{a_{n-1}}</math>. We have <math>a_1=x</math> and <math>a_2=2000</math>, and we compute: |
<cmath> | <cmath> | ||
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& = \frac{a_4+1}{a_3} | & = \frac{a_4+1}{a_3} | ||
= \frac{ \frac{2001 + x}{2000x} + 1 }{ \frac{2001}x } | = \frac{ \frac{2001 + x}{2000x} + 1 }{ \frac{2001}x } | ||
− | = \frac{ \frac{2001 + 2001x} }{ | + | = \frac{ \frac{2001 + 2001x}{2000x} }{ \frac{2001}x } |
= \frac{1+x}{2000} | = \frac{1+x}{2000} | ||
\\ | \\ | ||
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No two values of <math>x</math> we just computed are equal, and therefore there are <math>\boxed{4}</math> different values of <math>x</math> for which the sequence contains the value <math>2001</math>. | No two values of <math>x</math> we just computed are equal, and therefore there are <math>\boxed{4}</math> different values of <math>x</math> for which the sequence contains the value <math>2001</math>. | ||
− | |||
== See Also == | == See Also == | ||
− | {{AMC12 box|year=2001|num-b=| | + | {{AMC12 box|year=2001|num-b=24|after=Last question}} |
+ | {{MAA Notice}} |
Latest revision as of 12:53, 24 January 2024
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
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 (for all)
. This can be rewritten as
. We have
and
, and we compute:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.