Problem 1174

1. **Finding the number of trailing zeros in $2007!$:**

The number of trailing zeros in $2007!$ is determined by the number of times $10$ is a factor in $2007!$. Since $10 = 2 \times 5$, and there are always more factors of $2$ than $5$ in factorials, we only need to count the number of factors of $5$.

The number of factors of $5$ in $2007!$ can be found using the formula:
$$\left\lfloor \frac{2007}{5} \right\rfloor + \left\lfloor \frac{2007}{25} \right\rfloor + \left\lfloor \frac{2007}{125} \right\rfloor + \left\lfloor \frac{2007}{625} \right\rfloor$$

Calculating each term:
$$\left\lfloor \frac{2007}{5} \right\rfloor = \left\lfloor 401.4 \right\rfloor = 401$$
$$\left\lfloor \frac{2007}{25} \right\rfloor = \left\lfloor 80.28 \right\rfloor = 80$$
$$\left\lfloor \frac{2007}{125} \right\rfloor = \left\lfloor 16.056 \right\rfloor = 16$$
$$\left\lfloor \frac{2007}{625} \right\rfloor = \left\lfloor 3.2112 \right\rfloor = 3$$

Summing these values:
$$401 + 80 + 16 + 3 = 500$$

Therefore, the number of trailing zeros in $2007!$ is $500$.

2. **Finding the last non-zero digit of $2007!$:**

To find the last non-zero digit of $2007!$, we need to consider the factorial modulo $10$ while ignoring the factors of $10$ (i.e., factors of $2$ and $5$).

We use the fact that $4! \equiv -1 \mod 5$ and apply it repeatedly. This is based on the property of factorials and modular arithmetic.

First, we need to find $2007!$ modulo $5$:
$$2007! \equiv 2 \mod 5$$

This is because the product of numbers modulo $5$ cycles every $4$ numbers (since $4! \equiv -1 \mod 5$).

Next, we need to consider the powers of $2$ modulo $10$. Since $2007!$ has $500$ trailing zeros, we can ignore these factors of $10$ and focus on the remaining product modulo $10$.

By examining the pattern of the last non-zero digits of factorials, we find that the last non-zero digit of $2007!$ is $2$.

The final answer is $\boxed{ 500 }$ trailing zeros and the last non-zero digit is $2$.