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?
(In reply to
re: Paper and pencil solution (spoiler?) by Ady TZIDON)
I get 492 as well.
Basically the idea is that the trailing zeros of n! number one for each multiple of 5, plus an extra one for each multiple of a power of 5.
Since 625 can be analysed as 1*625, and 5*125, and 25*25, and 125*5, 625! has 156 trailing zeros. So 624! has 152.
Then 758 can be analysed as 1*625, and 6*125, and 30*25, and 151*5, so 758! has 188 trailing zeros.
When these factorials are multiplied together, the result will thus have 492 zeros.

Posted by broll
on 20140812 08:17:35 