604. Which of the regular polygons can completely and densely cover a plane without gaps?
604. Let there be a regular $n$-gon. As is known, the angle of this $n$-gon is $180^{\circ}-\frac{360^{\circ}}{n}$ or $180^{\circ}\left(1-\frac{2}{n}\right)$. It is required that at each vertex, the polygon, together with adjacent polygons, should tightly cover the plane. If $k$ regular polygons meet at one vertex, then the following equality must hold:

$$
k \cdot 180^{\circ}\left(1-\frac{2}{n}\right)=360^{\circ}
$$

or

$$
k=\frac{2}{1-\frac{2}{n}}=\frac{2 n}{n-2}
$$

We now need to determine such values for $\boldsymbol{n}(\boldsymbol{n}>2)$, so that $\boldsymbol{k}$ is an integer. The only values that satisfy this condition are 3, 4, and 6.

Indeed: $\frac{2 n}{n-2}=2+\frac{4}{n-2};$ for $n>6 \frac{4}{n-2}<1$, and therefore $k=\frac{2 n}{n-2}$ cannot be an integer. Consequently: the equilateral triangle, square, and regular hexagon are the only regular polygons that can tightly cover the plane.