2009 AIME I Problems/Problem 13

Revision as of 22:12, 20 March 2009 by Moplam (talk | contribs) (Solution)

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

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 $= 41+49=\boxed {090}$

See also

2009 AIME I (ProblemsAnswer KeyResources)
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