Nothing like finding a full explaination in a book! I just picked up Schaum's Outlines of College Mathematics (Ayres/Schmidt) and found the full explaination to this problem on page 132 - #20.3 ©. It is a Permutation-type problem (I stand corrected from an earlier posting) with a detailed discussion on how to solve problems like these.
Multiplicatin Principle applied:
First, we form all numbers ending in 0;
there are 6*5*1 of them.
Next, we form all numbers ending in 2, 4 or 6;
there are 5*5*3 of them.
(The 3 comes from the three even digits: 2,4,6)
Thus, in all, there are 6*5*1 + 5*5*3 = 105 numbers.
Hope this is a start to where you can find the answer...