Difference between revisions of "2009 AIME I Problems/Problem 13"
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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> | 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> | ||
+ | |||
+ | |||
+ | |||
+ | === 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 <math>a,b</math> such that <math>a_{1}=a_{3}</math> and <math>a_{2}=a_{4}.</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> 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}} | {{MAA Notice}} |
Revision as of 19:49, 7 September 2020
Problem
The terms of the sequence defined by for are positive integers. Find the minimum possible value of .
Solution
Solution 1
This question is guessable but let's prove our answer
lets put into now
and set them equal now
let's rewrite it
Let's make it look nice and let
Since and are integers, we can see is divisible by
But we can't have an infinite sequence of proper factors, unless
Thus,
So now, we know
To minimize , we need and
Thus, our answer
Solution 2
If , then either
or
All the integers between and would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.
So , which . When , . The smallest sum of two factors which have a product of is
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 such that and
Solving we find both equations end up to the equation in which we see to minimize we see that and 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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