$32 \cdot 10$ If $1+2+3+\cdots+k$ is equal to $N^{2}$, and $N<100$, then the possible values of $k$ are
(A) 1 .
(B) 1 and 8 .
(C) 8 .
(D) 8 and 49 .
(E) 1, 8 and 49 .
(7th American High School Mathematics Examination, 1956)
[Solution] From $1+2+3+\cdots+k=\frac{k(k+1)}{2}$, by the problem statement,
$$
\frac{k(k+1)}{2}=N^{2} \text {, i.e., } k^{2}+k-2 N^{2}=0 \text {. }
$$

When $k=1$, $\frac{1 \cdot 2}{2}=1=1^{2}$;
When $k=8$, $\frac{8 \cdot 9}{2}=36=6^{2}$;
When $k=49$, $\frac{49 \cdot 50}{2}=(7 \cdot 5)^{2}=35^{2}$.
Therefore, the answer is $(E)$.