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

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briggs@encompasserve.org

Posts: 404
Registered: 12/6/04
Re: Largest primeproduct with 2 as factor?
Posted: Aug 16, 2007 1:06 PM
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In article <1187278549.329963.136200@x35g2000prf.googlegroups.com>, Randy Poe <poespam-trap@yahoo.com> writes:
> 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).


2^4, 2^4-1 or 2^4-2 depending on whether you count all factors,
all factors other than 1, or all factors other than 1 and the number
itself.

[If you use the 2^n-2 interpretation then be sure to restrict your
attention to cases where n>0]

But that's assuming that p1, p2, p3 and p4 are all different.

If one or more are the same then you'd need to use a formula
for the number of subsets of a four element multiset.

If one pair is identical then it's 2*2*3 instead of 2^4.
If two pairs are identical then it's 3*3 instead of 2^4.
If there is one triplet then it's 2*4 instead of 2^4.
If all four are identical then it's 5 instead of 2^4.

In general, I believe that the number of subsets of a multiset is
the product of (one more than the multiplicity of the element) evaluated
over each distinct element in the multiset.

>> 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.


Or the full 2^8 if you count 1 as a factor.



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