(a) Calculate the sum:

$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{6}
$$

(b) Calculate the sum:

$$
\frac{1}{12}+\frac{1}{18}+\frac{1}{36}
$$

(c) Determine seven distinct positive integers such that the sum of their reciprocals is equal to 1.
Solution

(a) Reducing all fractions to a common denominator, we have:

$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{3+2+1}{6}=1
$$

(b) Again, we can reduce all fractions to a common denominator:

$$
\frac{1}{12}+\frac{1}{18}+\frac{1}{36}=\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{3+2+1}{36}=\frac{1}{6}
$$

(c) Note that using the two previous items, we can already add 5 reciprocals to get a sum of 1:

$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{18}+\frac{1}{36}=1
$$

To increase to 7 fractions, we just need to replace $\frac{1}{36}$ with the sum of the fractions $\frac{1}{2 \cdot 36}=\frac{1}{72}, \frac{1}{3 \cdot 36}=\frac{1}{108}$, and $\frac{1}{6 \cdot 36}=\frac{1}{216}$. We can make this replacement without changing the sum, since

$$
\frac{1}{72}+\frac{1}{108}+\frac{1}{216}=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right) \frac{1}{36}=\frac{1}{36}
$$

Thus, we can write that

$$
\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{18}+\frac{1}{72}+\frac{1}{108}+\frac{1}{216}=1
$$

Therefore, the seven integers are 2, 3, 12, 18, 72, 108, and 216.