Each of x, y and z are positive integers such that x+y+z=2006 and, x!*y!*z! = p*10q, 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*5q , 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
Paper and pencil solution (spoiler?)
