How to solve for boolean algebra?

Question

Which one of the following is the set of all atoms of $D_{78}$?
Select one:

a. $\{1,2,3\}$
b. $\{6,26,39\}$
c. $\{2,3,6\}$
d. $\{1,2,3,6\}$
e. $\{2,3,13\}$

Could you please explain why ‘e’ would be the correct answers, I am trying to understand posets and boolean algebra.

Note $D_{n}$ denotes the set of all positive integers which are divisors of $n$.

Answer 1

$78 = 2 \times 3 \times 13$ So $78$ is square free, which makes $D_{78}$ a Boolean algebra: It means that every divisor is of the form

$$d= 2^{a}\cdot 3^{b} \cdot 13^{c}, \text{ where } a,b,c \in \{0,1\}\text{.}$$

So this Boolean algebra is just isomorphic to $\{0,1\}^3$ as a product algebra, and its atoms are the divisors corresponding to the "unit vectors" $(1,0,0), (0,1,0), (0,0,1)$ corresponding to the prime divisors $2,3$ and $13$.

Answer 2

An atom is a minimal element in $D_{n}\setminus\{1\}$ (under divisibility, $1$ is the minimum element).

Thus it is an element that has no proper divisor except $1$. How do you call such numbers? Are there any in $D_{78}$?