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Topic: Largest primeproduct with 2 as factor?
Replies: 19   Last Post: Aug 16, 2007 1:06 PM

 Messages: [ Previous | Next ]
 Randy Poe Posts: 5,658 Registered: 12/6/04
Re: Largest primeproduct with 2 as factor?
Posted: Aug 16, 2007 11:35 AM

On Aug 16, 5:00 am, jonas.thornv...@hotmail.com wrote:
> On 16 Aug, 05:23, Randy Poe <poespam-t...@yahoo.com> wrote:
>
>
>

> > On Aug 15, 7:00 pm, jonas.thornv...@hotmail.com wrote:
>
> > > On 16 Aug, 00:42, Randy Poe <poespam-t...@yahoo.com> wrote:
>
> > > > On Aug 15, 6:36 pm, jonas.thornv...@hotmail.com wrote:
>
> > > > > On 16 Aug, 00:18, stephane.fr...@gmail.com wrote:
>
> > > > > > On Aug 15, 3:03 pm, jonas.thornv...@hotmail.com wrote:
>
> > > > > > > On 16 Aug, 00:00, stephane.fr...@gmail.com wrote:
>
> > > > > > > > On Aug 15, 2:53 pm, jonas.thornv...@hotmail.com wrote:
>
> > > > > > > > > I do not know much math, i just wonder if there is a largest
> > > > > > > > > primeproduct with two as one of the factors.
> > > > > > > > > Same i wonder for three.

>
> > > > > > > > > If not..., but if so is it true for any primefactor used in
> > > > > > > > > primeproduct that they will have a range?
> > > > > > > > > Do every factor have a range in primeproducts?

>
> > > > > > > > > A pure guess tell me that the numbers of primeproducts with 2 or 3
> > > > > > > > > used as one of the factor is infinite but i am not sure.

>
> > > > > > > > > JT
>
> > > > > > > > You guess is right. This is because there is infinity prime number
> > > > > > > > then if N is the bigger prime product, you will find A your biggest
> > > > > > > > prime with A = N/2. You can find a prime number A' with A'> A, then N'
> > > > > > > > = A x 2 > N. Same rules with 3 or any prime numbers.

>
> > > > > > > > SF
>
> > > > > > > Could anyone tell me the biggest "known" primeproduct that has two as
> > > > > > > factor?- Hide quoted text -

>
> > > > > > > - Show quoted text -
>
> > > > > > Well dude , you can suppose you got one and then prove that you can a
> > > > > > bigger one .- D?lj citerad text -

>
> > > > > > - Visa citerad text -
>
> > > > > Ok an easy question i do not have any primeproduct algorithm going, so
> > > > > i would be interested in a list of primeproducts less than a million
> > > > > that has two as factor?

>
> > > > > Can not be that many or?
>
> > > > > If not that many would be nice for 10 000 000 and for 100 000 000?
>
> > > > > Well if already a million would have couple of 100 i am not that
> > > > > interested in the bigger numbers.

>
> > > > What do you mean by "primeproduct"? Do you mean any
> > > > number that has 2 as a factor?

>
> > > > For instance, does 2^20 = 1048576 qualify as a "primeproduct"?
>
> > > > If you just mean any number that has two as a factor,
> > > > these are called "even numbers", and any number that
> > > > ends in 0, 2, 4, 6, or 8 is such a number. Thus, here
> > > > is a large even number: 7924720394820394812476

>
> > > Sorry i am tired, it is of course
> > > 6=2*3,10=2*5,14=2*7,22=2*11,26=2*13,34=2*17,38=2*19 and so son.
> > > and for three
> > > 6=3*2,15=3*5,21=3*7,33=3*11,39=3*13....

>
> > > So forget my question
>
> > Ah. Well, these days primes far > 10^6 are easily
> > turned out in milliseconds by computer software.

>
>
> > Not exactly what you're looking for, but if you
> > try various values of n you'll find a prime of the
> > size you want. Then obviously just multiply by 2.

>
> > For instance, the billionth prime is:
> > 22,801,763,489

>
> > - Randy- D?lj citerad text -
>
> > - Visa citerad text -
>
> If we multiply two composit (primeproducts), can they only have four
> factors?

To me, "composite" means any number which is not prime,
so 2*5*17*31*43*73 is a composite number, and obviously
it has more than four factors.

I think you are talking about numbers of the form
p1*p2 where p1 and p2 are prime. These do not have any
special name in English as far as I know. "Composite"
is the wrong word. I suppose we can define "primeproduct"
to mean such a number.

Given that definition, the product of two primeproducts
is a number of the form p1*p2*p3*p4. It has at most
four PRIME factors, but it also has as factors p1*p2,
p1*p3, p1*p4, p2*p4, p1*p2*p3, p1*p3*p4, etc. In
fact I think it has 2^4 - 1 factors (one for
every subset of {p1, p2, p3, p4} except the empty
set).

> And if we multiply two primeproducts of primeproducts can they only
> have eight factors?

Again, no. It has many more than 8 factors. I think
it has 2^8 - 1 factors.

- Randy