$11 \cdot 28$ in which base, $4 \cdot 13=100$?
(Kyiv Mathematical Olympiad, 1953)
[Solution] Let the base of the numeral system be $b$.
By the problem, it is clear that $b>4$.
At this point, we can obtain the equation
$$
\left\{\begin{array}{l}
4(1 \cdot b+3)=1 \cdot b^{2}+0 \cdot b+0, \\
b>4 .
\end{array}\right.
$$

Thus, we have
$$
\begin{array}{l}
b^{2}-4 b-12=0, \\
b=6, \quad b=-2 \text { (discard). }
\end{array}
$$

Therefore, the equation in the problem is in base 6.