Mrs. Teacher needed to come up with examples of equations for a written test. Therefore, she wrote down all equations of the form

$$
a \cdot x + b = 13
$$

where $a$ and $b$ are one-digit natural numbers. From all of them, she selected the equations whose root $x$ was 3. She put one equation into each group. What is the maximum number of groups there could be?

(K. Pazourek)
We know that $x=3$ is a solution to the given equation, so the equality

$$
a \cdot 3 + b = 13 \text{.}
$$

holds. For $a$ and $b$ to be natural numbers, $a$ must be either $1, 2, 3$, or $4$ (for $a=5$, we get $5 \cdot 3 = 15 > 13$ and $b$ would have to be negative, which is not possible). Now we substitute the individual values of $a$ into the equation and calculate the corresponding $b$:

- $a=1, b=10$,
- $a=2, b=7$,
- $a=3, b=4$,
- $a=4, b=1$.

We see that the case $a=1, b=10$ does not satisfy the conditions given in the problem. Therefore, there are exactly three pairs $(a, b)$ that solve the problem: $(2,7)$, $(3,4)$, and $(4,1)$. The teacher could thus form a maximum of three groups.

Grading. 1 point for substituting the root into the equation; 2 points for finding all three solutions; 3 points for explaining why there are no more solutions.