Difference between revisions of "2009 AIME I Problems/Problem 13"
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− | < | + | <cmath>a_{n + 2}(1 + a_{n + 1})= a_n + 2009</cmath> |
− | < | + | <cmath>a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009</cmath> |
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− | < | + | <cmath>a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009</cmath> |
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− | < | + | <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> |
− | < | + | <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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− | < | + | <cmath>(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n</cmath> |
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− | < | + | <cmath>(b_{n+1})(a_{n + 2}+1)= b_n</cmath> |
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− | < | + | <cmath>a_{n + 2}=a_n</cmath> |
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− | < | + | <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> |
Revision as of 23:14, 20 March 2009
Problem
The terms of the sequence defined by
for
are positive integers. Find the minimum possible value of
.
Solution
This question is guessable but let's prove our answer
let put into
now
and set them equal now
let's rewrite it
Let make it looks nice and let
Since and
are integer, 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
Thus, answer
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 |