3. (7p) Given the sets $A=\left\{x / x=11 a-3 ; a \in \mathbb{N}^{*}\right\}$ and $B=\{y / y=103-2 b ; b \in \mathbb{N}\}$. Determine $A \cap B$.

G.M. nr. 11/2015
Solution: $\quad A \cap B$ contains the common elements, so $x=y \Rightarrow 11a-3=103-2b \Rightarrow 11a+2b=106$. Since $2b$ and 106 are even numbers, it follows that $11a$ is also an even number, meaning $a$ is an even number. Since it is also non-zero, it follows that $a \in\{2,4,6,8\}$, for larger values the relation $11a+2b=106$ cannot be satisfied in $\mathbb{N}$. In conclusion, $A \cap B=\{19,41,63,85\}$.

## Scoring Guide

| $A \cap B$ contains the common elements, so $x=y \Rightarrow 11a-3=103-2b \Rightarrow 11a+2b=106$. | $2 p$ |
| :--- | :---: |
| Since $2b$ and 106 are even numbers, it follows that $11a$ is also an even number, meaning $a$ is an even number. | $2 p$ |
| Since it is also non-zero, it follows that $a \in\{2,4,6,8\}$, for larger values the relation $11a+2b=106$ cannot be satisfied. | $2 p$ |
| In conclusion, $A \cap B=\{19,41,63,85\}$. | $1 p$ |