Re: Maximum of Products
- From: David W. Cantrell <DWCantrell@xxxxxxxxxxx>
- Date: 22 May 2007 19:42:54 GMT
se16@xxxxxxxxxxxxxx wrote:
On 22 May, 08:00, isaac2004 <isaac_2...@xxxxxxxxx> wrote:
The first part is find the best sequenceIt is not incalculable as it is
with only integers. I found this to be a series of 500 2s. hence
2+2+2+2+2+2+....2+2+2=1000
2^500= ? which is an incalculable number,
3273390607896141870013189696827599152216642046043064789483291368096133796
4046745548832700923259041571508866841275600710092172565458853930533285275
89376 or about 3.27*10^150
But it is not the best sequence either.
Try 4+3+3+3+...+3+3=1000 or 3+3+3+...+3+3+2+2=1000 (with 332 3s) which
each give 4*3^332, i.e.
1014650697509413079627234489686710374303861808496103820156013382714388841
9190749328173921941991713634105804687612666026800024450257367553757158393
82921059431364 or about 1.01*10^159
In general for integers the best solution will be a number of 3s with
zero, one or two 2s (two 2s can be replaced by one 4).
Right.
It surprises me that I haven't seen anyone mention the solution when we do
not restrict the terms to being integers. Maybe someone did, and somehow I
overlooked it. Anyway, the best way is to use 368 terms of 1000/368. (Note
that rounding 1000/e gives 368.) The product is then
(1000/368)^368 = 5.8614...*10^159
Of course, that lies between the integer result above and
e^(1000/e) = 5.8615...*10^159 .
David
.
- References:
- Maximum of Products
- From: isaac2004
- Re: Maximum of Products
- From: se16
- Maximum of Products
- Prev by Date: Re: Normal subgroups of p-groups
- Next by Date: Re: Paths
- Previous by thread: Re: Maximum of Products
- Next by thread: Re: Maximum of Products
- Index(es):
Relevant Pages
|
