[Algebra]
Problem: A computer is programmed to choose an integer between 1 and 99, inclusive, so that the probability that it selects the integer $x$ is equal to $\log _{100}\left(1+\frac{1}{x}\right)$. Suppose that the probability that $81 \leq x \leq 99$ is equal to 2 times the probability that $x=n$ for some integer $n$. What is the value of $n$ ?

Solution: The probability that the integer $n$ is chosen is $\log _{100}\left(1+\frac{1}{n}\right)$.

The probability that an integer between 81 and 99 , inclusive, is chosen equals the sum of the probabilities that the integers $81,82, \ldots, 98,99$ are selected, which equals

$$
\log _{100}\left(1+\frac{1}{81}\right)+\log _{100}\left(1+\frac{1}{82}\right)+\cdots+\log _{100}\left(1+\frac{1}{98}\right)+\log _{100}\left(1+\frac{1}{99}\right)
$$

Since the second probability equals 2 times the first probability, the following equations are equivalent:

$$
\begin{aligned}
\log _{100}\left(1+\frac{1}{81}\right)+\log _{100}\left(1+\frac{1}{82}\right)+\cdots+\log _{100}\left(1+\frac{1}{98}\right)+\log _{100}\left(1+\frac{1}{99}\right) & =2 \log _{100}\left(1+\frac{1}{n}\right) \\
\log _{100}\left(\frac{82}{81}\right)+\log _{100}\left(\frac{83}{82}\right)+\cdots+\log _{100}\left(\frac{99}{98}\right)+\log _{100}\left(\frac{100}{99}\right) & =2 \log _{100}\left(1+\frac{1}{n}\right)
\end{aligned}
$$

Using logarithm laws, these equations are further equivalent to

$$
\begin{aligned}
\log _{100}\left(\frac{82}{81} \cdot \frac{83}{82} \cdots \cdot \frac{99}{98} \cdot \frac{100}{99}\right) & =\log _{100}\left(1+\frac{1}{n}\right)^{2} \\
\log _{100}\left(\frac{100}{81}\right) & =\log _{100}\left(1+\frac{1}{n}\right)^{2}
\end{aligned}
$$

Since logarithm functions are invertible, we obtain $\frac{100}{81}=\left(1+\frac{1}{n}\right)^{2}$.

Since $n>0$, then $1+\frac{1}{n}=\sqrt{\frac{100}{81}}=\frac{10}{9}$, and so $\frac{1}{n}=\frac{1}{9}$, which gives $n=9$.

Final answer: 9