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Difference between revisions of "2009 AIME I Problems/Problem 13"

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<cmath>\frac {2009}{a_{n+1}} < \frac {a_{n} + 2009}{1 + a_{n+1}} < \frac {a_{n}}{1} = a_{n}</cmath>
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<cmath>\frac {2009}{a_{n+1}} < \frac {2009 + a_{n}}{a_{n+1} + 1} < \frac {a_{n}}{1} = a_{n}</cmath>
  
 
All the integers between <math>a_{n}</math> and <math>\frac {2009}{a_{n+1}}</math> would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.
 
All the integers between <math>a_{n}</math> and <math>\frac {2009}{a_{n+1}}</math> would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.

Revision as of 20:47, 26 March 2009

Problem

The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$.

Solution

Solution 1

This question is guessable but let's prove our answer

\[a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}\]


\[a_{n + 2}(1 + a_{n + 1})= a_n + 2009\]


\[a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009\]


let put $n+1$ into $n$ now


\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009\]


and set them equal now


\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n\]


\[a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n\]


let's rewrite it


\[(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n\]


Let make it looks nice and let $b_n=a_{n + 2}-a_n$


\[(b_{n+1})(a_{n + 2}+1)= b_n\]


Since $b_n$ and $b_{n+1}$ are integer, we can see $b_{n+1}$ is divisible by $b_n$


But we can't have an infinite sequence of proper factors, unless $b_n=0$


Thus, $a_{n + 2}-a_n=0$


\[a_{n + 2}=a_n\]


So now, we know $a_3=a_1$


\[a_{3} = \frac {a_1 + 2009} {1 + a_{2}}\]


\[a_{1} = \frac {a_1 + 2009} {1 + a_{2}}\]


\[a_{1}+a_{1}a_{2} = a_1 + 2009\]


\[a_{1}a_{2} = 2009\]


To minimize $a_{1}+a_{2}$, we need $41$ and $49$


Thus, answer $= 41+49=\boxed {090}$

Solution 2

If $a_{n} \ne \frac {2009}{a_{n+1}}$, then either \[a_{n} = \frac {a_{n}}{1} < \frac {a_{n} + 2009}{1 + a_{n+1}} < \frac {2009}{a_{n+1}}\]

or

\[\frac {2009}{a_{n+1}} < \frac {2009 + a_{n}}{a_{n+1} + 1} < \frac {a_{n}}{1} = a_{n}\]

All the integers between $a_{n}$ and $\frac {2009}{a_{n+1}}$ would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.

So $a_{n} = \frac {2009}{a_{n+1}}$, which $a_{n} \cdot a_{n+1} = 2009$. When $n = 1$, $a_{1} \cdot a_{2} = 2009$. The smallest sum of two factors which have a product of $2009$ is $41 + 49=\boxed {090}$

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
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