Difference between revisions of "1998 AIME Problems/Problem 8"
m (→Solution 1) |
|||
Line 10: | Line 10: | ||
| 0 || 1 || 2 || 3 || 4 || 5 || 6 | | 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> | + | | <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>3000 - 5x</math> || <math>8x - 5000</math> |
|} | |} | ||
Line 17: | Line 17: | ||
Thus, | Thus, | ||
*<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> | + | *<math>n \equiv 1 \pmod{2}\quad\quad F_{n}\cdot 1000-F_{n-1}\cdot x</math> |
=== Solution 1 === | === Solution 1 === | ||
Line 42: | Line 42: | ||
=== Solution 2 === | === Solution 2 === | ||
− | It is well known that <math> | + | 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 == | == See also == | ||
Line 48: | Line 48: | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 18:38, 4 July 2013
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?
Contents
[hide]Solution
The best way to start is to just write out some terms.
0 | 1 | 2 | 3 | 4 | 5 | 6 |
aa | aaa | a |
By now its obvious that the numbers are related to the Fibonacci numbers.
Thus,
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.