04 May 05
Originally posted by PBE6Anayltically? I don't think there's a magic method of doing this, but it's easy to show rigorously that you've found all the solutions: all you need to do is remark that for all solutions in positive integers,
OK, here's what I got:
x = 1600; y = 8020
x = 1980; y = 158796
x = 2000; y = 802000
x = 2004; y = 0 is another pair, but 0 is not a positive integer.
Sadly, I solved this by trial and error. :'( If anyone knows how to do this analytically, I'd love to find out how.
1/x < 1/2005, so x < 2005,
and find y for each x in turn, confirming that the solutions you listed below are the only ones in positive integers.
1/2004 -1/0 is -infinity though, so that's not a solution at all.
04 May 05
Originally posted by AcolyteOops, my mistake (with the 0).
Anayltically? I don't think there's a magic method of doing this, but it's easy to show rigorously that you've found all the solutions: all you need to do is remark that for all solutions in positive integers,
1/x < 1/2005, so x < 2005,
and find y for each x in turn, confirming that the solutions you listed below are the only ones in positive integers.
1/2004 -1/0 is -infinity though, so that's not a solution at all.
04 May 05
2 edits
Originally posted by PBE6I think your 1600 should be 1604.
OK, here's what I got:
x = 1600; y = 8020
x = 1980; y = 158796
x = 2000; y = 802000
x = 2004; y = 0 is another pair, but 0 is not a positive integer.
Sadly, I solved this by trial and error. :'( If anyone knows how to do this analytically, I'd love to find out how.
x must be < 2005; so we need to find those values of x such that y = 2005x/(2005 - x) is also an integer.
Here's what I get for (x,y), and I think this is all such pairs:
(1604, 8020)
(1980, 158796)
(2000, 802000)
(2004, 4018020)
EDIT: I don't see any magical analytical solution either, although the problem can be whittled down quite a bit.
04 May 05
Originally posted by davegageOops again. I think my blood sugar is low. I need a cookie.
I think your 1600 should be 1604.
x must be < 2005; so we need to find those values of x such that y = 2005x/(2005 - x) is also an integer.
Here's what I get for (x,y), and I think this is all such pairs:
(1604, 8020)
(1980, 158796)
(2000, 802000)
(2004, 4018020)
EDIT: I don't see any magical analytical solution either, although the problem can be whittled down quite a bit.
Acolyte, bring me your head again.
04 May 05
Originally posted by PBE6Let a,b be relatively prime integers , and x=ka and y=kb, b>a
Oops again. I think my blood sugar is low. I need a cookie.
Acolyte, bring me your head again.
then k=c(b-a) where abc=2005 will give all the solutions:
x=(b-a)ca, y=(b-a)cb.
When c=1 there are two sets of factors to use for ab;
when c=5 or 401 there is just one set of factors for ab.
c cannot be 2005 as x=y not possible.
There are just these four since the only factors of 2005
are 1, 5, 401, and 2005.
[ 2004(1) , 2004(2005) ] , [ 396(5) , 396(401) ]
[400(5)(1) , 400(5)(401) ] , [ 4(401)(1) , 4(401)(5) ]