Peter, I appreciate your thoughtfulness. So the hasty reply may also be a wrong one. My apologies. From the example, it is obvious now that one has to consdier N1 and N9 simultaneously.

If anything comes up, I'll continue the thread. If anybody in your class has a solution, I'd like to see it.

>Actually i always thought i understood
>it easily, but then i
>got a really great black-out
>that made me wonder...

Good for you. I also thought it was pretty easy.

>P.S.: Was my reflection right that
>i can add the divisor
>1 to the inequation above,
>because it's a divisor you
>don't get by the product
>of N3 * N7 and
>it's in every integer?

Yes, 1 is one of the factor of any N and should be counted in N1.