Difference between revisions of "2005 AIME II Problems/Problem 11"
m (→See Also) |
|||
(13 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Let <math> | + | Let <math>m </math> be a positive integer, and let <math> a_0, a_1,\ldots,a_m </math> be a sequence of reals such that <math>a_0 = 37, a_1 = 72, a_m = 0, </math> and <math> a_{k+1} = a_{k-1} - \frac 3{a_k} </math> for <math> k = 1,2,\ldots, m-1. </math> Find <math>m. </math> |
− | + | ==Solution 1== | |
− | + | For <math>0 < k < m</math>, we have | |
− | |||
− | For <math> | ||
<center> | <center> | ||
− | <math> | + | <math>a_{k}a_{k+1} = a_{k-1}a_{k} - 3 </math>. |
</center> | </center> | ||
− | Thus the product <math> | + | Thus the product <math>a_{k}a_{k+1}</math> is a [[monovariant]]: it decreases by 3 each time <math>k</math> increases by 1. For <math>k = 0</math> we have <math>a_{k}a_{k+1} = 37\cdot 72</math>, so when <math>k = \frac{37 \cdot 72}{3} = 888</math>, <math>a_{k}a_{k+1}</math> will be zero for the first time, which implies that <math>m = \boxed{889}</math>, our answer. |
+ | |||
+ | |||
+ | Note: In order for <math>a_{m} = 0</math> we need <math>a_{m-2}a_{m-1}=3</math> simply by the recursion definition. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Plugging in <math>k = m-1</math> to the given relation, we get <math>0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}</math>. Inspecting the value of <math>a_{k}a_{k+1}</math> for small values of <math>k</math>, we see that <math>a_{k}a_{k+1} = 37\cdot 72 - 3k</math>. Setting the RHS of this equation equal to <math>3</math>, we find that <math>m</math> must be <math> \boxed{889}</math>. | ||
+ | |||
+ | ~ anellipticcurveoverq | ||
+ | |||
+ | ==Induction Proof== | ||
+ | As above, we experiment with some values of <math>a_{k}a_{k+1}</math>, conjecturing that <math>a_{m-p}a_{m-p-1}</math> = <math>3p</math> ,where <math>m</math> is a positive integer and so is <math>p</math>, and we prove this formally using induction. The base case is for <math>p = 1</math>, <math>a_{m} = a_{m-2} - 3/a_{m-1}</math> Since <math>a_{m} = 0 </math>, <math>a_{m-1}a_{m-2} = 3</math>; from the recursion given in the problem <math>a_{m-p+1} = a_{m-p-1} - 3/a_{m-p}</math>, so <math>a_{m-p+1} = 3p/a_{m-p} - 3/a_{m-p} = 3(p-1)/a_{m-p}</math>, so <math>a_{m-p}a_{m-p+1} = a_{m-(p-1)}a_{m-(p-1)-1} = 3(p-1)</math>, hence proving our formula by induction. | ||
+ | ~USAMO2023 | ||
+ | |||
+ | ==Solution 3 (Telescoping)== | ||
+ | Note that <math> a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3 </math>. Then, we can generate a sum of series of equations such that <math>\sum_{k=1}^{m-1} a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3(m-1)</math>. Then, note that all but the first and last terms on the LHS cancel out, leaving us with <math>a_m\cdot a_{m-1} - a_1\cdot a_0 = -3(m-1)</math>. Plugging in <math>a_m=0</math>, <math>a_0=37</math>, <math>a_1=72</math>, we have <math>-37\cdot 72 = -3(m-1)</math>. Solving for <math>m</math> gives <math>m=\boxed{889}</math>. | ||
+ | ~sigma | ||
+ | |||
+ | |||
+ | ==Video solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=JfxNr7lv7iQ | ||
== See also == | == See also == | ||
Line 19: | Line 39: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:52, 21 November 2022
Contents
Problem
Let be a positive integer, and let be a sequence of reals such that and for Find
Solution 1
For , we have
.
Thus the product is a monovariant: it decreases by 3 each time increases by 1. For we have , so when , will be zero for the first time, which implies that , our answer.
Note: In order for we need simply by the recursion definition.
Solution 2
Plugging in to the given relation, we get . Inspecting the value of for small values of , we see that . Setting the RHS of this equation equal to , we find that must be .
~ anellipticcurveoverq
Induction Proof
As above, we experiment with some values of , conjecturing that = ,where is a positive integer and so is , and we prove this formally using induction. The base case is for , Since , ; from the recursion given in the problem , so , so , hence proving our formula by induction. ~USAMO2023
Solution 3 (Telescoping)
Note that . Then, we can generate a sum of series of equations such that . Then, note that all but the first and last terms on the LHS cancel out, leaving us with . Plugging in , , , we have . Solving for gives . ~sigma
Video solution
https://www.youtube.com/watch?v=JfxNr7lv7iQ
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.