
Re: Michael's conjecture
Posted:
Aug 6, 2013 3:47 PM


"0" <marty.musatov@gmail.com> wrote in message news:e2d2470d2871486996a62d25546c53ca@googlegroups.com... > 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 nonprimes, e.g. 6 divides 12 and 4 divides 12, but 24 does not.
Regards, Mike.

