Difference between revisions of "1998 AIME Problems/Problem 8"
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*<math>n \equiv 0 \pmod{2}\quad\quad F_{n-1}\cdot 1000-F_n\cdot x</math> | *<math>n \equiv 0 \pmod{2}\quad\quad F_{n-1}\cdot 1000-F_n\cdot x</math> | ||
| − | *<math>n \equiv 1 \pmod{2}\quad\quad | + | *<math>n \equiv 1 \pmod{2}\quad\quad F_{n}\cdot x-F_{n-1}\cdot 1000</math> |
where the <math> F_{n}</math> are [[Fibonacci number]]s. This can be proven through induction. | where the <math> F_{n}</math> are [[Fibonacci number]]s. This can be proven through induction. | ||
Latest revision as of 22:33, 28 March 2025
Problem
Except for the first two terms, each term of the sequence 
is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encounted. What positive integer 
produces a sequence of maximum length?
Solution
The best way to start is to just write out some terms.
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
 aa |
 aaa |
 |
 a |
 |
 |
|
It is now apparent that each term can be written as
where the 
are Fibonacci numbers. This can be proven through induction.
Solution 1
We can start to write out some of the inequalities now:
And in general,
It is apparent that the bounds are slowly closing in on , so we can just calculate for some large value of 
(randomly, 10, 11):
Thus the sequence is maximized when 
Solution 2
It is well known that 
, so 
approaches
See also
| 1998 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
