How many zeros are at the end of (100!)(200!)(300!) when multiplied out?
The number of zeros at the end of a number is equivalent to the number of factors of 10 that number has. Since there are more factors of 2 than there are of 5 in a factorial, this is determined by the number of factors of 5. So, we go about computing this for each factorial separately.

To count the number of zeroes at the end of $100!$, we must count the number of factors of 5 in the product.  There are $\left\lfloor \frac{100}{5}\right\rfloor$ multiples of 5 from 1 to 100. (The notation $\left\lfloor x\right\rfloor$ means the greatest integer less than or equal to $x$, so basically, $\left\lfloor \frac{100}{5}\right\rfloor$ means "divide 100 by 5 and round down.")  This gives us 20 multiples of 5.  But the multiples of 25 contribute an additional factor of 5, so we have to add in the total number of multiples of 25, which gives us a total number of factors of 5 of $\left\lfloor \frac{100}{5} \right\rfloor + \left\lfloor \frac{100}{25} \right\rfloor = 20+ 4 = 24$.

Similarly, for $200!$ the contributed zeros total $\left\lfloor \frac{200}{5} \right\rfloor + \left\lfloor \frac{200}{25} \right\rfloor + \left\lfloor \frac{200}{125} \right\rfloor = 40 + 8 + 1 = 49$; and for $300!$, $\left\lfloor \frac{300}{5} \right\rfloor + \left\lfloor \frac{300}{25} \right\rfloor + \left\lfloor \frac{300}{125} \right\rfloor = 60 + 12 + 2 = 74$. So, our answer is $24 + 49 + 74 = \boxed{147}$.