2009 AIME I Problems/Problem 13

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 $41 + 49=\boxed {090}$

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
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2009 AIME I Problems/Problem 13

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