Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1998 AIME Problems/Problem 8"

(→‎Solution 2)
(→‎Solution)
Line 3: Line 3:
  
 
__TOC__
 
__TOC__
−
== Solution ==
 
−
The best way to start is to just write out some terms.
 
−
 
−
{| class="wikitable"
 
−
|-
 
−
| 0 || 1 || 2 || 3 || 4 || 5 || 6
 
−
|- 
 
−
| <math>\quad 1000 \quad</math><font color="white">aa</font> || <math>\quad x \quad</math><font color="white">aaa</font> || <math>1000 - x</math> || <math>2x - 1000</math><font color="white">a</font> || <math>2000 - 3x</math> || <math>5x - 3000</math> || <math>5000 - 8x</math>
 
−
|}
 
−
 
−
It is now apparent that each term can be written as
 
−
 
−
*<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 F_{n}\cdot 1000-F_{n-1}\cdot x</math>
 
−
 
−
where the <math> F_{n}</math> are [[Fibonacci number]]s. This can be proven through induction.
 
−
 
−
=== Solution 1 ===
 
−
We can start to write out some of the inequalities now:
 
−
 
−
#<math>x > 0</math>
 
−
#<math>1000 - x > 0 \Longrightarrow x < 1000</math>
 
−
#<math>2x - 1000 > 0 \Longrightarrow x > 500</math>
 
−
#<math>2000 - 3x > 0 \Longrightarrow x < 666.\overline{6}</math>
 
−
#<math>5x - 3000 > 0 \Longrightarrow x > 600</math>
 
−
 
−
And in general,
 
−
 
−
*<math>n \equiv 0 \pmod{2}\quad\quad x < \frac{F_{n-1}}{F_n} \cdot 1000</math>
 
−
*<math>n \equiv 1 \pmod{2}\quad\quad x > \frac{F_{n-1}}{F_{n}} \cdot 1000</math>
 
−
 
−
It is apparent that the bounds are slowly closing in on <math>x</math>, so we can just calculate <math>x</math> for some large value of <math>n</math> (randomly, 10, 11):
 
−
 
−
<math>x < \frac{F_{7}}{F_{8}} \cdot 1000 = \frac{13}{21} \cdot 1000 = 618.\overline{18}</math>
 
−
 
−
<math>x > \frac{F_{8}}{F_{9}} \cdot 1000 = \frac{21}{34} \cdot 1000 \approx 617.977</math>
 
−
 
−
Thus the sequence is maximized when <math>x = \boxed{618}.</math>
 
−
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1998|num-b=7|num-a=9}}
 
{{AIME box|year=1998|num-b=7|num-a=9}}

Revision as of 01:00, 25 August 2024

Problem

Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ 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 $x$ produces a sequence of maximum length?

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_8&oldid=225731"