5. There are 15 identical chocolate bars, each of which can be divided into at most two smaller pieces (which may not be equal). If these 15 chocolate bars can be evenly distributed among $n$ students, for the following 5 values: $7, 11, 13, 16, 20$, the number of values that $n$ can take is ( ).
(A) 5
(B) 4
(C) 3
(D) 2
5.A.

Consider each piece of chocolate as 1 unit.
When $n=7$, 4 pieces of chocolate are each divided into $\frac{1}{7}$ and $\frac{6}{7}$, 4 pieces of chocolate are each divided into $\frac{2}{7}$ and $\frac{5}{7}$, 4 pieces of chocolate are each divided into $\frac{3}{7}$ and $\frac{4}{7}$, and another 3 pieces are not divided. Then
$$
\begin{aligned}
15= & 4\left(\frac{1}{7}+\frac{6}{7}\right)+4\left(\frac{2}{7}+\frac{5}{7}\right)+ \\
& 4\left(\frac{3}{7}+\frac{4}{7}\right)+3 \times 1 \\
= & \left(1+4 \times \frac{1}{7}+2 \times \frac{2}{7}\right)+2\left(1+2 \times \frac{4}{7}\right)+ \\
& 2\left(\frac{3}{7}+2 \times \frac{6}{7}\right)+\left(2 \times \frac{2}{7}+2 \times \frac{3}{7}+\frac{5}{7}\right)+ \\
& 3 \times \frac{5}{7} .
\end{aligned}
$$

When $n=11$, 2 pieces are each divided into $\frac{k}{11}$ and $\frac{11-k}{11} (k=1,2,3,4,5)$, and another 5 pieces are not divided.

When $n=13$, 2 pieces are each divided into $\frac{k}{13}$ and $\frac{13-k}{13} (k=1,2, \cdots, 6)$, and another 3 pieces are not divided.
When $n=16$, each piece is divided into $\frac{1}{16}$ and $\frac{15}{16}$.
When $n=20$, each piece is divided into $\frac{5}{20}$ and $\frac{15}{20}$.