4. We are looking at four-digit PIN codes. We say that a PIN code dominates another PIN code if each digit of the first PIN code is at least as large as the corresponding digit of the second PIN code. For example, 4961 dominates 0761, because $4 \geqslant 0, 9 \geqslant 7, 6 \geqslant 6$ and $1 \geqslant 1$. We want to assign a color to every possible PIN code from 0000 to 9999, but if one PIN code dominates another, they must not have the same color.

What is the minimum number of colors with which you can do this?
B4. 37 To get an idea of how many colors we need, we will make a list of pin codes where each subsequent pin code dominates all previous pin codes. We start at 0000 and end at 9999. For example, you can find the following list:

$$
0000,0001,0002, \ldots, 0009,0019, \ldots, 0099,0199, \ldots, 0999,1999, \ldots, 9999
$$

There are many other possible lists from 0000 to 9999. All pin codes in this list must receive a different color. For this list, you already need 37 colors.

We will now show that 37 colors are sufficient to color all possible pin codes. If we add up all four digits of a pin code, we get a number which we call the strength of the pin code. Pin code 0000 has strength 0 and 9999 has strength 36. All pin codes have a strength between 0 and 36. If pin code $A$ is dominated by another pin code, say $B$, then the strength of $B$ is greater than the strength of $A$. (The converse is not true: if pin code $B$ has a greater strength than pin code $A$, it does not necessarily mean that pin code $B$ dominates pin code $A$. For example, consider 2021 with strength 5 and 0089 with strength 17: 2021 is not dominated by 0089.)

We will use the strength to color the pin codes. We use one color per strength. Thus, all pin codes with the same strength receive the same color, and pin codes with different strengths always receive different colors. In this way, it can never happen that pin code $A$ is dominated by another pin code $B$, but $A$ and $B$ have the same color. The strength of a pin code is a number from $0,1, \ldots, 36$. With 37 colors, we can thus color all pin codes in the required way. We had already seen that at least 37 colors are needed, so this is also the minimum number of colors you need.