PROBLEM 2. Let the set $A=\left\{n \mid n=2^{a} \cdot 3^{b} \cdot 5^{c}\right.$, where $\left.a, b, c \in \mathbb{N}\right\}$.

a) Show that $81 \in A$;

b) Determine the elements of the set $A$ that have exactly 6 divisors.
PROBLEM 2. Let the set $A=\left\{n \mid n=2^{a} \cdot 3^{b} \cdot 5^{c}\right.$, where $\left.a, b, c \in \mathbb{N}\right\}$.

a) Show that $81 \in A$;

b) Determine the elements of the set $A$ that have exactly 6 divisors.

## Grading Rubric.

(4p) a) From $81=2^{0} \cdot 3^{4} \cdot 5^{0}$, it follows that $81 \in A$.

$(1 p) b)$ Since the number of divisors of a number of the form $n=2^{a} \cdot 3^{b} \cdot 5^{c}$ is $(a+1)(b+1)(c+1)$,

(2p) we have the following cases:

| $a$ | $b$ | $c$ | $n$ |
| :---: | :---: | :---: | :---: |
| 0 | 0 | 5 | 3125 |
| 0 | 5 | 0 | 243 |
| 5 | 0 | 0 | 32 |
| 0 | 1 | 2 | 75 |
| 0 | 2 | 1 | 45 |
| 1 | 2 | 0 | 18 |
| 1 | 0 | 2 | 50 |
| 2 | 1 | 0 | 12 |
| 2 | 0 | 1 | 20 |