sorry, of course i should have given an example.

it's easy to understand that the product of every pair of factors, of which one ends in 3 and the other in 7, produces a factor that ends in 1.

so you mentioned the following example:

>Let's also try 4641 = 3·7·13·17. 4641 has the following >factors: 1, 3, 7, 13, 17, 21, 39, 51, 91, 119, 221, 273, 357, >663, 1547, 4641.

so there are 4 divisors ending in 3 (3, 13, 273, 663), so

N3 = 4,

and other 4 divisors end in 7 (7, 17, 357, 1547), so

N7 = 4. Consequently N3 * N7 = 16, but N1 = 6.

But for example the pair of divisors 663 and 1547 doesn't produce another divisor ending in 1, as their product produces a higher number than the considered integer itself.

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

thank you very much,
peter

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?