# Number theory

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I seek a proof, (using the properties of the h.c.f.),  that if k divides nr then k = k1k2  where k1 divides n and k2 divides r

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Use fndamental theorem of arithmetic.

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• 2 months later...
On 04/05/2018 at 8:39 PM, caledonia said:

I seek a proof, (using the properties of the h.c.f.),  that if k divides nr then k = k1k2  where k1 divides n and k2 divides r

Well, as a layman, it would seem that if [k] = [k1k2], then [k] =  itself, as  *  = . This is just speculation, let's get more intense?

If the [k] = prime and divides into [k2], then six would be the lowest figure of the range, as the lowest figure for [k1] would be , as the lowest prime, so, we could run a log of  to find the 'ratios' of one to two, of course. this would follow, basically, that if [k1] = , then [k2] = , and k = , and that [k1] would actually be [k/3], and, that [k2] would be [k/2].

So, if [k=18] as [k=3=18] then [n] would be [k/3].

But, I am just a layman in this vast world.

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