Difference between revisions of "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

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

$$a_{n + 2}(1 + a_{n + 1})= a_n + 2009$$

$$a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009$$

lets 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's make it look 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 integers, we can see $b_n$ is divisible by $b_{n+1}$

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

Solution 3 (BS Solution)

Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms.

\begin{align*} a_{1} &= a \\ a_{2} &= b \\ a_{3} &= \displaystyle\frac{a+2009}{1+b} \\ a_{4} &= \displaystyle\frac{(b+1)(b+2009)}{a+b+2010} \\ \end{align*}

The terms get more and more wacky, so we just solve for $a,b$ such that $a_{1}=a_{3}$ and $a_{2}=a_{4}.$

Solving we find both equations end up to the equation $ab=2009$ in which we see to minimize we see that $a = 49$ and $b=41.$ This solution is VERY non rigorous and not recommended.

 2009 AIME I (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions