Difference between revisions of "2009 AIME I Problems/Problem 13"
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| − | < | + | <cmath>a_{n + 2}(1 + a_{n + 1})= a_n + 2009</cmath> |
| − | < | + | <cmath>a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009</cmath> |
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| − | < | + | <cmath>a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009</cmath> |
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| − | < | + | <cmath>a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n</cmath> |
| − | < | + | <cmath>a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n</cmath> |
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| − | < | + | <cmath>(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n</cmath> |
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| − | < | + | <cmath>(b_{n+1})(a_{n + 2}+1)= b_n</cmath> |
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| − | < | + | <cmath>a_{n + 2}=a_n</cmath> |
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| − | < | + | <cmath>a_{3} = \frac {a_1 + 2009} {1 + a_{2}}</cmath> |
| − | < | + | <cmath>a_{1} = \frac {a_1 + 2009} {1 + a_{2}}</cmath> |
| − | < | + | <cmath>a_{1}+a_{1}a_{2} = a_1 + 2009</cmath> |
| − | < | + | <cmath>a_{1}a_{2} = 2009</cmath> |
Revision as of 23:14, 20 March 2009
Problem
The terms of the sequence 
defined by 
for 
are positive integers. Find the minimum possible value of 
.
Solution
This question is guessable but let's prove our answer
![\[a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}\]](http://latex.artofproblemsolving.com/4/a/4/4a4433c819327e5380119b8a6ecbb30f8ecb3029.png)
![\[a_{n + 2}(1 + a_{n + 1})= a_n + 2009\]](http://latex.artofproblemsolving.com/a/b/2/ab25f1150bcd8353ea5e24f6ad7d99f860365a7a.png)
![\[a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009\]](http://latex.artofproblemsolving.com/c/1/8/c1852d4c3b5c6d2b2f15d7f6aacaee7148bf48b3.png)
let put 
into 
now
![\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009\]](http://latex.artofproblemsolving.com/6/f/d/6fd094e3590c265542bc7fef6dcae93468213305.png)
and set them equal now
![\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n\]](http://latex.artofproblemsolving.com/f/1/6/f165a7ebacf76fb0a49dd560f0f3ea3100e5ae6e.png)
![\[a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n\]](http://latex.artofproblemsolving.com/f/8/8/f88df2a0d334ff7f23264fb59721a30706b310cc.png)
let's rewrite it
![\[(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n\]](http://latex.artofproblemsolving.com/e/1/d/e1deaa74007757c90386bf522a73064fd1482900.png)
Let make it looks nice and let 
![\[(b_{n+1})(a_{n + 2}+1)= b_n\]](http://latex.artofproblemsolving.com/9/b/0/9b0ba0198502a99c37ced38bff90e8ef8f522f48.png)
Since 
and 
are integer, we can see is divisible by
But we can't have an infinite sequence of proper factors, unless 
Thus, 
![\[a_{n + 2}=a_n\]](http://latex.artofproblemsolving.com/5/4/5/54507b2b35e52fa27d2953f7d8c12bffefaff930.png)
So now, we know 
![\[a_{3} = \frac {a_1 + 2009} {1 + a_{2}}\]](http://latex.artofproblemsolving.com/f/2/6/f26554de5ca84fa268767fe51a9f57ab57486d7f.png)
![\[a_{1} = \frac {a_1 + 2009} {1 + a_{2}}\]](http://latex.artofproblemsolving.com/e/c/5/ec5631c4fdec6e4031b1f0b08e49055b3117b2f6.png)
![\[a_{1}+a_{1}a_{2} = a_1 + 2009\]](http://latex.artofproblemsolving.com/7/7/7/777abcb534a891a2b25bede47961e842230193fd.png)
![\[a_{1}a_{2} = 2009\]](http://latex.artofproblemsolving.com/7/0/c/70c3dd4f18868927aebeaedd56442d91192bc238.png)
To minimize 
, we need 
Thus, answer 
See also
| 2009 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
