Difference between revisions of "1985 AIME Problems/Problem 13"
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== See also == | == See also == | ||
− | * [[ | + | {{AIME box|year=1985|num-b=12|num-a=14}} |
− | * [[ | + | * [[AIME Problems and Solutions]] |
− | * [[ | + | * [[American Invitational Mathematics Examination]] |
− | + | * [[Mathematics competition resources]] | |
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] |
Revision as of 14:36, 6 May 2007
Problem
The numbers in the sequence ,
,
,
,
are of the form
, where
For each
, let
be the greatest common divisor of
and
. Find the maximum value of
as
ranges through the positive integers.
Solution
If denotes the greatest common divisor of
and
, then we have
. Now assuming that
divides
, it must divide
if it is going to divide the entire expression
.
Thus the equation turns into . Now note that since
is odd for integral
, we can multiply the left integer,
, by a multiple of two without affecting the greatest common divisor. Since the
term is quite restrictive, let's multiply by
so that we can get a
in there.
So . It simplified the way we wanted it to!
Now using similar techniques we can write
. Thus
must divide
for every single
. This means the largest possible value for
is
, and we see that it can be achieved when
.
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |