Difference between revisions of "1993 USAMO Problems/Problem 4"
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Let <math>a</math>, <math>b</math> be odd positive integers. Define the sequence <math>(f_n)</math> by putting <math>f_1 = a</math>, | Let <math>a</math>, <math>b</math> be odd positive integers. Define the sequence <math>(f_n)</math> by putting <math>f_1 = a</math>, | ||
− | <math>f_2 = b</math>, and by letting | + | <math>f_2 = b</math>, and by letting <math>f_n</math> for <math>n\ge3</math> be the greatest odd divisor of <math>f_{n-1} + f_{n-2}</math>. |
Show that <math>f_n</math> is constant for <math>n</math> sufficiently large and determine the eventual | Show that <math>f_n</math> is constant for <math>n</math> sufficiently large and determine the eventual | ||
value as a function of <math>a</math> and <math>b</math>. | value as a function of <math>a</math> and <math>b</math>. | ||
− | |||
== Solution == | == Solution == | ||
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Then, <math>f_{n+1}\le\frac{f_{n}+f_{n-1}}{2}< \max (f_{n},f_{n-1})</math>, <math>f_{n+2}\le\frac{f_{n+1}+f_{n}}{2}< \max (f_{n},f_{n+1})\le\max (f_{n},f_{n-1})</math> | Then, <math>f_{n+1}\le\frac{f_{n}+f_{n-1}}{2}< \max (f_{n},f_{n-1})</math>, <math>f_{n+2}\le\frac{f_{n+1}+f_{n}}{2}< \max (f_{n},f_{n+1})\le\max (f_{n},f_{n-1})</math> | ||
− | Thus, <math>\max (f_{n}, | + | Thus, <math>\max (f_{n},f_{n-1})> \max (f_{n+1},f_{n+2})</math> and that means that <math>\max (f_{2n},f_{2n+1})</math> is a strictly decreasing function and it must reach <math>0</math> as <math>n\rightarrow\infty</math>, which contradict with the fact that <math>f_n>0</math>. |
<br/><b>Part 1</b> proven. | <br/><b>Part 1</b> proven. | ||
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<P align="right"><math>\mathbb{Q.E.D}</math></P> | <P align="right"><math>\mathbb{Q.E.D}</math></P> | ||
− | == | + | == See Also == |
{{USAMO box|year=1993|num-b=3|num-a=5}} | {{USAMO box|year=1993|num-b=3|num-a=5}} | ||
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks] | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 07:56, 19 July 2016
Problem 4
Let ,
be odd positive integers. Define the sequence
by putting
,
, and by letting
for
be the greatest odd divisor of
.
Show that
is constant for
sufficiently large and determine the eventual
value as a function of
and
.
Solution
Part 1) Prove that is constant for sufficiently large
.
Note that if there is some for any
, then
, which is odd. Thus,
and by induction, all
is constant for
.
Also note that since average of
positive number is always positive.
Thus, assume for contradiction, ,
.
Then, ,
Thus, and that means that
is a strictly decreasing function and it must reach
as
, which contradict with the fact that
.
Part 1 proven.
Part 2) Show that the constant is .
For any where
.
for
with the same property except with
and
.
Therefore, if I prove that the constant for any with relatively prime
,
is
, then I have shown that part 2 is true.
Lemma) If , then
.
Assume for contradiction that , since both
and
are odd,
is not divisible by
.
for some
such that
is odd.
, where
and
is another integer.
Thus, is divisible by
which contradicts with the assumption that
.
Lemma proven
By induction, since
.
Since there must exist some where
(part 1),
.
See Also
1993 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.