Difference between revisions of "2004 AIME II Problems/Problem 9"
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Let <math>x = a_2</math>. It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what <math>x</math> can be. Finding that <math>x = 5</math> works, after bashing out the rest of the terms we find that <math>a_{16} = 957</math> and <math>a_{17} = 1089</math>, hence our answer is <math>957 + 16 = \boxed{973}</math>. | Let <math>x = a_2</math>. It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what <math>x</math> can be. Finding that <math>x = 5</math> works, after bashing out the rest of the terms we find that <math>a_{16} = 957</math> and <math>a_{17} = 1089</math>, hence our answer is <math>957 + 16 = \boxed{973}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | We can find the value of <math>a_{9}</math> by its bounds using three conditions: | ||
+ | |||
+ | |||
+ | #<math>0<a_{8} = 2a_{9}-a_{10}</math> | ||
+ | #<math>a_{10} < a_{11}</math> (note that the sequence must be increasing on all terms, not monotonically increasing) <math>a_{10} < \frac{a_{10}^2}{a_{9}} \rightarrow a_{9} < a_{10}</math> | ||
+ | #<math>a_{11} = \frac{a_{10}^2}{a_{9}} = \frac{(646-a_{9})^2}{a_{9}}</math>, so necessarily <math>a_{9}</math> is a factor of <math>646^2</math>, which factorizes to <math>2^2\cdot 17^2 \cdot 19^2</math> | ||
+ | |||
+ | Rearranging conditions 1 and 2, we get: | ||
+ | |||
+ | <cmath>\frac{646}{3} < a_{9} < \frac{646}{2}</cmath> | ||
+ | |||
+ | trying all the terms from the third condition, it is clear that <math>a_9 = 289</math> is the only solution. | ||
+ | Then we can calculate the next few terms from there since we have <math>a_{10}</math> as well, to find that <math>a_{16} = 957</math> and <math>a_{17} = 1089</math>, thus we have our answer of <math>957 + 16 = \boxed{973}</math>. | ||
+ | |||
+ | ~KafkaTamura | ||
== See also == | == See also == | ||
+ | * https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_10 | ||
{{AIME box|year=2004|n=II|num-b=8|num-a=10}} | {{AIME box|year=2004|n=II|num-b=8|num-a=10}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:41, 7 August 2024
Problem
A sequence of positive integers with and
is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all
the terms
are in geometric progression, and the terms
and
are in arithmetic progression. Let
be the greatest term in this sequence that is less than
. Find
Solution 1
Let ; then solving for the next several terms, we find that
, and in general,
,
, where
.[1]
From , we find that by either the quadratic formula or trial-and-error/modular arithmetic that
. Thus
, and we need to find the largest
such that either
. This happens with
, and this is the
th term of the sequence.
The answer is .
^ We can show this by simultaneous induction: since
and
Solution 2
Let . It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what
can be. Finding that
works, after bashing out the rest of the terms we find that
and
, hence our answer is
.
Solution 3
We can find the value of by its bounds using three conditions:
(note that the sequence must be increasing on all terms, not monotonically increasing)
, so necessarily
is a factor of
, which factorizes to
Rearranging conditions 1 and 2, we get:
trying all the terms from the third condition, it is clear that is the only solution.
Then we can calculate the next few terms from there since we have
as well, to find that
and
, thus we have our answer of
.
~KafkaTamura
See also
2004 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.