Problem: Suppose $a$ $b$ and $c$ are positive integers with $a+b+c=2006$ , and $a!b!c!=m\cdot 10^n$ , where $m$ and $n$ are integers and $m$ is not divisible by $10$ . What is the smallest possible value of $n$
$\mathrm{(A)}\ 489 \qquad \mathrm{(B)}\ 492  \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 498 \qquad \mathrm{(E)}\ 501$

Solution: Clearly, the power of $2$ that divides $n!$ is larger or equal than the power of $5$ which divides
it. Hence we are trying to minimize the power of $5$ that will divide $a!b!c!$
Consider $n! = 1\cdot 2 \cdot \dots \cdot n$ . Each fifth term is divisible by $5$ , each $25$ -th one
by $25$ , and so on. Hence the total power of $5$ that divides $n$ is $\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \cdots$ . (For any $n$ only finitely many terms in the sum
are
non-zero.)
In our case we have $a<2006$ , so the largest power of $5$ that will be less than $a$ is at most $5^4 = 625$ . Therefore the power of $5$ that divides $a!$ is equal to $\left\lfloor \frac a{5}\right\rfloor + \left\lfloor \frac a{25}\right\rfloor + \left\lfloor \frac a{125}\right\rfloor + \left\lfloor \frac a{625}\right\rfloor$ . The same
is true for $b$ and $c$
Intuition may now try to lure us to split $2006$ into $a+b+c$ as evenly as possible, giving $a=b=669$ and $c=668$ . However, this solution is not optimal.
To see how we can do better, let's rearrange the terms as follows:
\begin{align*} \text{result} & = \Big\lfloor \frac a{5}\Big\rfloor + \Big\lfloor \frac b{5}\Big\rfloor + \Big\lfloor \frac c{5}\Big\rfloor \\ & + \Big\lfloor \frac a{25}\Big\rfloor + \Big\lfloor \frac b{25}\Big\rfloor + \Big\lfloor \frac c{25}\Big\rfloor \\ & + \Big\lfloor \frac a{125}\Big\rfloor + \Big\lfloor \frac b{125}\Big\rfloor + \Big\lfloor \frac c{125}\Big\rfloor \\ & + \Big\lfloor \frac a{625}\Big\rfloor + \Big\lfloor \frac b{625}\Big\rfloor + \Big\lfloor \frac c{625}\Big\rfloor \end{align*}
The idea is that the rows of the above equation are roughly equal to $\left\lfloor \frac n{5}\right\rfloor$ $\left\lfloor \frac n{25}\right\rfloor$ , etc.
More precisely, we can now notice that for any positive integers $a,b,c,k$ we can write $a,b,c$ in the form $a=a_0k + a_1$ $b=b_0k+b_1$ $c=c_0k + c_1$ , where all $a_i,b_i,c_i$ are integers and $0\leq a_1,b_1,c_1<k$
It follows that \[\Big\lfloor \frac a{k}\Big\rfloor + \Big\lfloor \frac b{k}\Big\rfloor + \Big\lfloor \frac c{k}\Big\rfloor = a_0+b_0+c_0\] and \[\Big\lfloor \frac {a+b+c}k\Big\rfloor = a_0 + b_0 + c_0 + \Big\lfloor \frac {a_1+b_1+c_1}k\Big\rfloor \leq a_0 + b_0 + c_0 + 2\]
Hence we get that for any positive integers $a,b,c,k$ we have \[\Big\lfloor \frac a{k}\Big\rfloor + \Big\lfloor \frac b{k}\Big\rfloor + \Big\lfloor \frac c{k}\Big\rfloor \quad \geq \quad \Big\lfloor \frac {a+b+c}k\Big\rfloor - 2\]
Therefore for any $a,b,c$ the result is at least $\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \left\lfloor \frac n{625}\right\rfloor - 8 = 401 + 80 + 16 + 3 - 8 = 500 - 8 = 492$
If we now show how to pick $a,b,c$ so that we'll get the result $492$ , we will be done.
Consider the row with $625$ in the denominator. We need to achieve sum $1$ in this row,
hence we need to make two of the numbers smaller than $625$ . Choosing $a=b=624$ does this, and it will give us the largest possible remainders for $a$ and $b$ in
the other three rows, so this is a pretty good candidate. We can compute $c=2006-a-b=758$ and verify that this triple gives the desired result $\boxed{492}$

Answer:  492