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Topic: Michael's conjecture
Replies: 16   Last Post: Aug 7, 2013 12:55 PM

 Messages: [ Previous | Next ]
 Mike Terry Posts: 767 Registered: 12/6/04
Re: Michael's conjecture
Posted: Aug 6, 2013 3:47 PM

"0" <marty.musatov@gmail.com> wrote in message
> Thanks. I was interested in the play of integers with three factors. Such
as a number having 2, 3, 5 as factors also being divisible by 6, 15, as well
as 30. Can you comment?

That's correct. Integers have an essentially unique factorisation into
primes, and if you look at the prime factorisation for an integer which is
divisible by 2, 3, and 5, you will see 2, 3 and 5 appearing, e.g. like this:

840 = 2 * 2 * 2 * 3 * 5 * 7
^ ^ ^

(840 is divisible by primes 2,3,5 and we naturally see 2,3,5 in the prime
factorisation.)

Clearly by inspection the various products of these primes are also going to
divide 840, just by rearranging the prime factorisation, e.g.:

840 = (2 * 3 * 5) * (2 * 2 * 7) = 30 * 28, so 30 divides 840 as we would
expect.

The key result underlying this is the essentially unique factorisation of
integers into primes, which as Rupert points out was a theorem known to the
Ancient Greeks...

Note this argument doesn't work with non-primes, e.g. 6 divides 12 and 4
divides 12, but 24 does not.

Regards,
Mike.

Date Subject Author
8/5/13
8/5/13
8/5/13 Rupert
8/6/13 Virgil
8/6/13
8/6/13
8/6/13 Mike Terry
8/6/13
8/6/13
8/6/13 quasi
8/6/13
8/6/13
8/6/13 Don Redmond
8/7/13 Don Redmond
8/7/13 Michael Klemm
8/7/13
8/7/13