Zeroes all around

$290=\left \lfloor \frac{n}{5} \right \rfloor+ \left \lfloor \frac{n}{5^2} \right \rfloor+ \left \lfloor \frac{n}{5^3} \right \rfloor+\ldots$

$290=\frac{\frac{n}{5}}{1-\frac{1}{5}}$

$290=\frac{n}{4}\Rightarrow n=1160$

Now we can see that number of zeroes in $1160!=288$ .

Therefore to get two more zeroes (i.e. $290$ zeroes) the answer must be $\boxed{1170}$ .

We will have to evaluate the range of possible value of $n$ , using $\sum_{i=1}^{\left \lfloor log_5n \right \rfloor} \left \lfloor \frac{n}{5^i} \right \rfloor$ For $n = 1100$ , these sum will be $220+44+8+1 = 273$ and for $n = 1200$ , these sum will be $240+48+9+1 = 298$ . Given we need $n!$ with at least 290 trailing zeros, we can conclude that $1100 < n < 1200$ . When we shift $n = 1150$ , the sum will be $230+46+9+1 = 286$ . From here, every increment of 5 will add one more zero. We then add four 5s (=20), then we will get the sum of $234+46+9+1 = 290$ , satisfying the requirement, hence, $n=1170$