$p$ is a prime number such that the period of its decimal reciprocal is 200 . That is,

$$
\frac{1}{p}=0 . X X X X \ldots
$$

for some block of 200 digits $X$, but

$$
\frac{1}{p} \neq 0 . Y Y Y Y \ldots
$$

for all blocks $Y$ with less than 200 digits. Find the 101st digit, counting from the left, of $X$.
Let $X$ be a block of $n$ digits and let $a=0 . X \ldots$ Then $10^{n} a=X . X \ldots$. Subtracting the previous two equalities gives us $\left(10^{n}-1\right) a=X$, i.e. $a=\frac{X}{10^{n}-1}$.

Then the condition that $a=\frac{1}{p}$ reduces to $\frac{1}{p}=\frac{X}{10^{n}-1}$ or $p X=10^{n}-1$. For a given $p$ and $n$, such an $X$ exists if and only if $p$ divides $10^{n}-1$. Thus $p$ divides $10^{200}-1$ but not $10^{n}-1,1 \leq n \leq 199$. Note that $10^{200}-1$ can be factored in this way:

$$
\begin{aligned}
10^{200}-1 & =\left(10^{100}\right)^{2}-1 \\
& =\left(10^{100}-1\right)\left(10^{100}+1\right) .
\end{aligned}
$$

Since $p$ is prime and does not divide $10^{100}-1$, it must divide $10^{100}+1$, so that $10^{100}+1=k p$ for an integer $k$ and $X=\frac{\left(10^{100}-1\right)\left(10^{100}+1\right)}{p}=\left(10^{100}-1\right) k$.

If $p=2,3,5$, or 7 , the fraction $\frac{1}{p}$ either terminates or repeats less than 200 digits. Therefore $p>10$ and $k<\frac{10^{100}}{p}<10^{99}$. Now let us calculate the 101st digit of $X=10^{100} k-k$, i.e. the digit representing multiples of $10^{99}$. Since $10^{100} k$ is divisible by $10^{100}$, its $10^{99} \mathrm{~s}$ digit and all later digits are 0 . Since $k<10^{99}$, $k$ does not contribute a digit to the $10^{99} \mathrm{~s}$ place, but it generates a borrow to this place, changing it into a 9. Thus the 101st digit of $X$ is a 9 .