Each of x, y and z are positive integers such that x+y+z=2006 and, x!*y!*z! = p*10^{q}, where each of p and q is a positive integer and p is not divisible by 10.

What is the smallest value of q?

The problem might as well say

x!*y!*z! = p*5^{q} , because there are more than enough 2's for each factor of 5. The objective, I think is to minimize the total factors of 625, and then the total factors of 125 that are not a multiple of 625, and then the total factors of 25 that are not a multiple of 125, and then the total factors of 5 that are not a multiple of 25.

If two of the numbers are right below 625, I think that is one way of achieving the minimum. I am thinking about 624 + 624 + 758.

If I have counted right,

5! is a multiple of 5^1

25! is a multiple of 5^6 (because 5*1+ 1 = 6)

125! is a multiple of 5^31 (because 5*6+ 1 = 31)

625! is a multiple of 5^156 (because 5*31 + 1 = 156)

so 624! is a multiple of 5^152

758! is a multiple of 5^184 (ie, 156 + 31 - 3)

So the minimum q, I think is 152 + 152 + 184 = **488**

Of course, I may have made a mistake ...

/**********************************/

Yes, I did make a mistake!

758! is a multiple of 5^**188** (ie, 156 + 31 **+1**),

**because 758 = 625 + 125 + 8**

So the minimum q, is 152 + 152 + **188** = **492**

*Edited on ***August 12, 2014, 10:52 am**