$36 \cdot 19$ At the end of a banquet, hands were shaken 28 times in total. Assuming each person at the banquet treated all other attendees with the same courtesy, the number of people attending the banquet was
(A) 14.
(B) 28.
(C) 56.
(D) 8.
(E) 7.
(2nd American High School Mathematics Examination, 1951)
[Solution] If the number of attendees is $n$, then the number of handshakes is $C_{n}^{2}=\frac{1}{2} n(n-1)$. From
$$
\frac{1}{2} n(n-1)=28 \text {, }
$$

we get
$$
n^{2}-n-56=0 \text {. }
$$

Thus,
$n=8$ or -7 (discard).

Therefore, the answer is $(D)$.