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

Latest revision as of 02:01, 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?

Solutions were removed

@@ Line 2: / Line 2: @@
 Except for the first two terms, each term of the sequence <math>1000, x, 1000 - x,\ldots</math> 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 <math>x</math> produces a sequence of maximum length?
-__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>
-|}
-By now its obvious that the numbers are related to the [[Fibonacci number]]s.
-Thus,
+Solutions were removed
-*<math>n \equiv 0 \pmod{2}\quad\quad F_{n-1}\cdot 1000-F_n\cdot x</math>
+__TOC__
-*<math>n \equiv 1 \pmod{2}\quad\quad F_{n}\cdot 1000-F_{n-1}\cdot x</math>
-=== 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>3000 - 2x > 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_{9}}{F_{10}} \cdot 1000 = \frac{34}{55} \cdot 1000 = 618.\overline{18}</math>
-<math>x > \frac{F_{10}}{F_{11}} \cdot 1000 = \frac{55}{89} \cdot 1000 \approx 617.977</math>
-Thus the sequence is maximized when <math>x = 618</math>.
-=== Solution 2 ===
-It is well known that <math>\lim_{n\rightarrow\infty} \frac{F_{n-1}}{F_n} = \phi - 1 =\frac{1 + \sqrt{5}}{2} - 1 \approx .61803</math>, so <math>1000 \cdot \frac{F_{n-1}}{F_n}</math> approaches <math>x = 618</math>.
 == 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

Page

Toolbox

Search

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

Latest revision as of 02:01, 25 August 2024

Problem

Contents

See also