[Number Theory]
Problem: Compute the sum of the reciprocals of the positive integer divisors of 24.

Solution: The map $n \mapsto 24 / n$ establishes a one-to-one correspondence among the positive integer divisors of 24 . Thus

$$
\begin{aligned}
\sum_{\substack{n \mid 24 \\
n>0}} \frac{1}{n} & =\sum_{\substack{n \mid 24 \\
n>0}} \frac{1}{24 / n} \\
& =\frac{1}{24} \sum_{\substack{n \mid 24 \\
n>0}} n
\end{aligned}
$$

Because $24=2^{3} \cdot 3$, the sum of the positive divisors of 24 is $\left(1+2+2^{2}+2^{3}\right)(1+3)=15 \cdot 4=60$. Hence the sum is $60 / 24=\mathbf{5} / \mathbf{2}$.
Because $24=2^{3} \cdot 3$, any positive divisor of 24 is of the form $2^{a} 3^{b}$ where $a=0,1,2$, or 3 , and $b=0$ or 1 . So the sum of the positive divisors of 24 can be represented as the product $(1+2+4+8)(1+3)$. Similarly, the sum of their reciprocals can be represented as the product $\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)\left(\frac{1}{1}+\frac{1}{3}\right)$. The first sum is $\frac{15}{8}$ and the second is $\frac{4}{3}$, so the product is $\mathbf{5 / 2}$.

Final answer: $\frac{5}{2}$