Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Integer pairs in sum of reciprocals
Replies: 39   Last Post: Jan 22, 2001 6:02 PM

 Messages: [ Previous | Next ]
 355113@my-deja.com Posts: 19 Registered: 12/13/04
Re: Integer pairs in sum of reciprocals
Posted: Jan 11, 2001 12:08 PM

In article <3A5139E4.94731FAA@hot.rr.com>,
"Ross A. Finlayson" <rfinlayson@hot.rr.com> wrote:
>
> saxon970@yahoo.com wrote:

> >
> > How many pairs of positive integers a and b are there such that
> > a < b and 1/a + 1/b = 1/2001 ?

>
> Set a to 2001 or greater and b to a large value. The result is
> very close to 1/2001, where 2001 is called c,
>
> 1/a + 1/b = 1/c, or
>
> a ^-1 + b ^-1 = c^-1
>
> There are probably infinite solutions.

saxon's message has disappeared from Deja, but as several
people mentioned (including Nathan until he changed his mind!)
there are 13 solutions:

Writing 1/a + 1/b = 1/2001 as u.v = 2001^2 = 3^2.23^2.29^2,
with u, v = a - 2001, b - 2001, there are 3^3 = 27 integer
pairs |u|, |v|.

(Each 3 here is for the 3 possible powers 0, 1, 2 of each
prime independently that can divide say u. More generally,
if the number was p^m.q^n.. with p, q,.. prime then the
number of pairs would be (m+1).(n+1)..).

If u, v < 0 then for 0 < a, b we require |u|, |v| < 2001
which is impossible. Thus 0 < u, v.

There are thus 13 such pairs with 0 < u < v, i.e. half
of the 26 pairs with u not equal v.

Cheers

John_Ramsden@acer.de

Sent via Deja.com
http://www.deja.com/