AIME 1998 · 第 8 题

Solution

The best way to start is to just write out some terms.

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6

$\quad 1000 \quad$aa

$\quad x \quad$aaa

$1000 - x$

$2x - 1000$a

$2000 - 3x$ $5x - 3000$ $5000 - 8x$

It is now apparent that each term can be written as

• $n \equiv 0 \pmod{2}\quad\quad F_{n-1}\cdot 1000-F_n\cdot x$
• $n \equiv 1 \pmod{2}\quad\quad F_{n}\cdot x-F_{n-1}\cdot 1000$

where the $F_{n}$ are Fibonacci numbers. This can be proven through induction.

Solution 1

We can start to write out some of the inequalities now:

1. $x > 0$
2. $1000 - x > 0 \Longrightarrow x < 1000$
3. $2x - 1000 > 0 \Longrightarrow x > 500$
4. $2000 - 3x > 0 \Longrightarrow x < 666.\overline{6}$
5. $5x - 3000 > 0 \Longrightarrow x > 600$

And in general,

• $n \equiv 0 \pmod{2}\quad\quad x < \frac{F_{n-1}}{F_n} \cdot 1000$
• $n \equiv 1 \pmod{2}\quad\quad x > \frac{F_{n-1}}{F_{n}} \cdot 1000$

It is apparent that the bounds are slowly closing in on $x$, so we can just calculate $x$ for some large value of $n$ (randomly, 10, 11):

$x < \frac{F_{9}}{F_{10}} \cdot 1000 = \frac{34}{55} \cdot 1000 = 618.\overline{18}$ $x > \frac{F_{10}}{F_{11}} \cdot 1000 = \frac{55}{89} \cdot 1000 \approx 617.977$

Thus the sequence is maximized when $x = \boxed{618}.$

Solution 2

It is well known that $\lim_{n\rightarrow\infty} \frac{F_{n-1}}{F_n} = \phi - 1 =\frac{1 + \sqrt{5}}{2} - 1 \approx .61803$, so $1000 \cdot \frac{F_{n-1}}{F_n}$ approaches $x = \boxed{618}.$