2009 AIME I Problems/Problem 13
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.
$\begin{align*}
a_{1} &= a \\
a_{2} &= b \\
a_{3} &=\frac{a+2009}{1+b} \\
a_{4} &=\frac{(b+1)(b+2009)}{a+b+2010} \\
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
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