To determine how many zeros are at the end of $200!$, we need to count the number of factors of 10 in $200!$. Since each factor of 10 is the product of a factor of 2 and a factor of 5, the number of trailing zeros is determined by the number of factors of 5 (because there are always more factors of 2 than factors of 5 in factorials).

The number of factors of 5 in $200!$ can be found by summing:

$\left\lfloor \frac{200}{5} \right\rfloor + \left\lfloor \frac{200}{25} \right\rfloor + \left\lfloor \frac{200}{125} \right\rfloor$

This counts how many multiples of 5, 25, and 125 there are in $200!$.

Now let's calculate each term:

$\left\lfloor \frac{200}{5} \right\rfloor = 40, \quad \left\lfloor \frac{200}{25} \right\rfloor = 8, \quad \left\lfloor \frac{200}{125} \right\rfloor = 1$

Adding these together:

$40 + 8 + 1 = 49$

Therefore, the number of trailing zeros in $200!$ is 49.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. How many zeros are at the end of $100!$?
2. What is the prime factorization of $200!$?
3. How does the number of trailing zeros change for factorials of powers of 10 (e.g., $1000!$)?
4. How would the number of trailing zeros change if we considered base 8 instead of base 10?
5. How many digits are there in $200!$?

Tip: The formula used to count trailing zeros can be generalized to other bases by considering the prime factors of the base.