5 . Find the number of divisors of 1984, as well as the sum of all its divisors.
Let the number of divisors of 1984 be $\mathrm{n}$, and the sum of all divisors be $S$. Since $1984=2^{8} \cdot 31$, the following product
$$
\left(1+2+2^{2}+\cdots+2^{8}\right)(1+31)
$$

the terms of the expanded form of this product are all divisors of 1984, and 1984 has no other divisors. Therefore, the number of divisors is equal to the number of terms in the expanded form of this product. The sum of all divisors is equal to this product, that is
$$
\begin{array}{l}
\mathrm{n}=(6+1)(1+1)=14, \\
\mathrm{~S}=\frac{2^{7}-1}{2-1} \cdot \frac{31^{2}-1}{31-1}=4064 .
\end{array}
$$

In fact, these 14 divisors are $1, 31, 2, 2^{2}, 2^{3}$,
$$
\begin{array}{l}
4, 2^{5}, 2^{8}, 2 \cdot 31, 2^{2} \cdot 31, 2^{3} \cdot 31, 2^{4} \cdot 31, 2^{5} \cdot 31 \\
2^{0} \cdot 31
\end{array}
$$