What is the remainder when $1995! -1$ is divided by $100 * 101$? What is the remainder when $1995! - 1$ is divided by $100 * 101$?
 A: First observe that
$$1995! = 1 \times 2 \times \cdots \times 100 \times 101 \times \cdots \times 1995$$
so that 
$$1995! \equiv 0 \pmod{100 \cdot 101}$$
which implies that
$$1995! -1 \equiv -1 \equiv 10099 \pmod{100 \cdot 101}$$

EDIT:
The above solution does not actually require knowledge on modular arithmetic. You can think of it like this:
First observe that
$$1995! = 1 \times 2 \times \cdots \times 100 \times 101 \times \cdots \times 1995 = 100 \times 101 \times \text{some integer}$$
It follows that $1995!$ is a multiple of $100 \times 101$.
Hence, $1995!-1$ leaves a remainder of $100 \times 101 -1 = 10099$ when divided by $100 \times 101$
A: Note that 1995! is divisible by 101*100. Then 1995!-10100 is also divisible by 10100. No number between 1995! and 1995!-10100 is divisible by 10100. Therefore the remainder will be 1995!-1-(1995!-10100)=10099
A: Let $n_{1}$ and $n_{2}$ ....$n_r$ be a set of integers that are less than $x$.
Then, in this case $n_1$.$n_2$.$n_3$......$n_r$ would always divide $x!$ leaving the remainder zero.
Hence, $x!-q$ where $q$ is less than $n_1.n_2.n_3....n_r$ would always leave a remainder {${n_1.n_2....n_r}$ - $q$} when $x!-q$ is divided by it.
Hence, $1995! - 1$, where $1$ is less than $100$ X $101$ and $100$ and $101$ are both less than $1995$, would leave a remainder $100$ X $101$ - $1$, i.e. $10099$ when divided by $100$ X $101$.
