1. ## Minimization problem

Supose $a,b,c$ are positive integers so that $a+b+c=2010$ and $a!*b!*c!=m*10^n$ where $m$ and $n$ are integers and $m$ is not divisible by 10. Which is the minimum value $n$ can have?

The first thing I did was to find the values a,b,c should have to minimize the multiplication of factorials. Dividing 2010 by 3 = 670, then a=669, b=670 and c=671. Now I stuck, can someone help?

2. If $p$ is a prime and $t=fcnt(n,p)=\sum_{k=1}^{\infty} \left[\frac{n}{p^k}\right]$, then $n!=p^t M, \quad p \not | M$
If $p_1 < p_2$, then $fcnt(n,p_1) \geq fcnt(n,p_2)$, thus $n! = M 10^{fcnt(n,5)}$
For example, $fct(671,5) = 134+26+5+1 = 166$ and so $671! = a 10^{166}$ where a is not divisible by 10

I think you can use this to solve the problem