10. On the table, there are 1989 identically sized paper pieces placed without overlapping. How many different colors are needed at minimum to color each paper piece so that, regardless of how these paper pieces are placed, it is always possible to color them such that any two touching paper pieces are of different colors?
10. Solution: At least 4 different colors are needed, otherwise, if only 1, 2, 3 three colors are used, (you can construct the graph yourself) assume that among the 6 paper pieces on the left, one of $A, B$ is the first color, the other is the second color, and $C$ is a different color from $B$ and not the third color, so $C$ and $A$ are the same color. This way, $D, E$ are both the third color, which leads to a contradiction! Therefore, 3 colors are not enough.

Next, change 1989 to $n$, and use induction on $n$ to prove: the paper circles can be colored with 4 colors such that any two tangent circles have different colors.
(1) For $n=1, 2, 3$, it is obviously true.
(2) Assume it is true for $n-1$ ($n \geqslant 4$), for $n$, consider the convex hull of the centers of these $n$ circles, let $A$ be one of the vertices of this convex hull, then the paper circle $T$ centered at $A$ is tangent to at most 3 other paper circles. By the induction hypothesis, all the paper circles except $T$ can be colored with 4 colors, then color $T$ with a color different from the colors of the circles tangent to $T$. This way, it is also true for $n$.