A new problem for 9-12 might be:
The "five-torial" of a number is the normal factorial, except each factor of five has multiplicity 2. For instance, 7 five-torial is 7*6*5*5*4*3*2*1.
In a "five five-torial" each factor of five has multiplicity of 3.
What is the minimum number of fives in a five...five-torial such that the last non-zero digit of 2013 five....five-torial is odd?
-Sean