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

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The terms of the sequence <math>(a_i)</math> defined by <math>a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}</math> for <math>n \ge 1</math> are positive integers. Find the minimum possible value of <math>a_1 + a_2</math>.
 
The terms of the sequence <math>(a_i)</math> defined by <math>a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}</math> for <math>n \ge 1</math> are positive integers. Find the minimum possible value of <math>a_1 + a_2</math>.
  
−
== Solution ==
+
==Solution 1==
−
<math>a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}</math>
 
  
 +
This question is guessable but let's prove our answer
  
−
<math>a_{n + 2}(1 + a_{n + 1})= a_n + 2009</math>
+
<cmath>a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}</cmath>
  
  
−
<math>a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009</math>
+
<cmath>a_{n + 2}(1 + a_{n + 1})= a_n + 2009</cmath>
  
  
−
let put <math>n+1</math> into <math>n</math> now
+
<cmath>a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009</cmath>
  
  
−
<math>a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009</math>
+
lets put <math>n+1</math> into <math>n</math> now
 +
 
 +
 
 +
<cmath>a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009</cmath>
  
  
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−
<math>a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n</math>
+
<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>
  
  
−
<math>a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n</math>
+
<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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<math>(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n</math>
+
<cmath>(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n</cmath>
  
  
−
Let make it looks nice and let <math>b_n=a_{n + 2}-a_n</math>
+
Let's make it look nice and let <math>b_n=a_{n + 2}-a_n</math>
  
  
−
<math>(b_{n+1})(a_{n + 2}+1)= b_n</math>
+
<cmath>(b_{n+1})(a_{n + 2}+1)= b_n</cmath>
  
  
−
Since <math>b_n</math> and <math>b_{n+1}</math> are integer, we can see <math>b_{n+1}</math> is divisible by <math>b_n</math>
+
Since <math>b_n</math> and <math>b_{n+1}</math> are integers, we can see <math>b_n</math> is divisible by <math>b_{n+1}</math>
  
  
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<math>a_{n + 2}=a_n</math>
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<cmath>a_{n + 2}=a_n</cmath>
  
  
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<math>a_{3} = \frac {a_1 + 2009} {1 + a_{2}}</math>
+
<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>
 +
 
 +
 
 +
To minimize <math>a_{1}+a_{2}</math>, we need <math>41</math> and <math>49</math>
 +
 
 +
 
 +
Thus, our answer <math>= 41+49=\boxed {090}</math>
 +
 
 +
==Solution 2==
  
 +
If <math>a_{n} \ne \frac {2009}{a_{n+1}}</math>, then either
 +
<cmath>a_{n} = \frac {a_{n}}{1} < \frac {a_{n} + 2009}{1 + a_{n+1}} < \frac {2009}{a_{n+1}}</cmath>
  
−
<math>a_{1} = \frac {a_1 + 2009} {1 + a_{2}}</math>
+
or
  
 +
<cmath>\frac {2009}{a_{n+1}} < \frac {2009 + a_{n}}{a_{n+1} + 1} < \frac {a_{n}}{1} = a_{n}</cmath>
  
−
<math>a_{1}+a_{1}a_{2} = a_1 + 2009</math>
+
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.
  
 +
So <math>a_{n} = \frac {2009}{a_{n+1}}</math>, which <math>a_{n} \cdot a_{n+1} = 2009</math>. When <math>n = 1</math>, <math>a_{1} \cdot a_{2} = 2009</math>. The smallest sum of two factors which have a product of <math>2009</math> is <math>41 + 49=\boxed {090}</math>
  
−
<math>a_{1}a_{2} = 2009</math>
 
  
  
−
To minimize <math>a_{1}+a_{2}</math>, we need <math>41 and 49</math>
+
== 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.
 +
<cmath>
 +
\begin{align*}
 +
a_{1} &= a \\
 +
a_{2} &= b \\
 +
a_{3} &=\frac{a+2009}{1+b} \\
 +
a_{4} &=\frac{(b+1)(b+2009)}{a+b+2010} \\
 +
\end{align*}
 +
</cmath>
 +
The terms get more and more wacky, so we just solve for <math>a,b</math> such that <math>a_{1}=a_{3}</math> and <math>a_{2}=a_{4}.</math>
  
−
Thus, answer <math>= 41+49=\boxed {090}</math>
+
Solving we find both equations end up to the equation <math>ab=2009</math> in which we see to minimize we see that <math>a = 49</math> and <math>b=41</math> or vice versa for an answer of <math>\boxed{90}.</math> This solution is VERY non rigorous and not recommended.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2009|n=I|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 15:39, 15 February 2021

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 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} &=\frac{a+2009}{1+b} \\ a_{4} &=\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 $b=41$ or vice versa for an answer of $\boxed{90}.$ This solution is VERY non rigorous and not recommended.

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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