and of course I too omitted part of the whole proof. That proofs my point at how easy it is to not close up a proof !

Everything was covered in my last post except for the numbers 2's (or 5's) that can not be taken up into a combination (2*5=10). Can this excess in number 2's (or 5's) result in trailing zero's ?

No, as all the numbers that can be fully factorized into a multiplication of 2's is a closed number space (ring) as well. Multiplication using numbers that are members of this number space (ring) all have a last digit that belongs to the set 2,4,6 and 8. Therefor there exist no multiplication of any given number of 2's that can produce a number that ends with the digit 0. Therefor this number can never be evenly divided by 10. As the set 2,4,6 and 8 is a subset of the set 2,3,4,6,7,8 and 9 the numbers space (ring) for powers of 2's is wholely enclosed in the numbers space (ring) for the leftovers as identified in my former posting. As a result, the multiplication of the leftovers from factorizing and the excess of 2's can never produce a number that ends with either a 5 or 0 as the last digit. Therefor this series of leftover numbers can never produce additional trailing zero's.

The proof when an excess of 5's is had goes along the line of reasoning.

So ! Now the proof is mathematically complete and closed.


Can you tell that I do this kind of stuff more often !


Ehh, is someone still paying attention ?! <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" />

Wouter