2. a) (3p) Let $A=\left\{\frac{2024}{13}, \frac{2025}{14}, \frac{2026}{15}, \ldots\right\}$ and $B=\{x \in \mathbb{N} \mid x \in A\}$. Find the cardinality of the set $B$.

Supliment Gazeta Matematică, April 2014

b) (4p) Determine the numbers $a, b, c \in \mathbb{N}$, with $a<b$ such that $\frac{a^{2}+a}{2}+\frac{b^{2}+b}{2}=\frac{c+3}{c+1}$.

Gazeta Matematică nr. 1/2014
Solution: a) The set $A$ can also be written in the form

$A=\left\{\frac{2011+13}{13}, \frac{2011+14}{14}, \frac{2011+15}{15}, \ldots\right\}=\left\{\frac{2011}{13}+1, \frac{2011}{14}+1, \frac{2011}{15}+1, \ldots\right\}$. Since $B$ is formed by the natural numbers in $A$, which are of the form $\frac{2011}{n}+1$, it follows that $n$ must be a divisor of 2011 greater than or equal to 13. Since 2011 is a prime number, it follows that the set $B$ has only one element $\frac{2011}{2011}+1=2$; card $B=1$.
b) The given relation can be written as $\frac{a(a+1)}{2}+\frac{b(b+1)}{2}=1+\frac{2}{c+1}$. But $a(a+1)$ and $b(b+1)$ are products of consecutive numbers $\Rightarrow \frac{2}{c+1} \in \mathbb{N}$, hence $c \in\{0 ; 1\}$. For $c=0$ we have $a(a+1)+b(b+1)=6$, with the variants $a=0 ; b=2$ and $a=2 ; b=0$. For $c=1$ we have $a(a+1)+b(b+1)=4$, with $a=1 ; b=1$. But $a<b$, so the only solution is $a=0 ; b=2 ; c=0$.

## Grading:

a) $A=\left\{\frac{2011+13}{13}, \frac{2011+14}{14}, \frac{2011+15}{15}, \ldots\right\}=\left\{\frac{2011}{13}+1, \frac{2011}{14}+1, \frac{2011}{15}+1, \ldots\right\}$

Since $B$ is formed by the natural numbers in $A$, which are of the form $\frac{2011}{n}+1$, it follows that $n$ must be a divisor of 2011 greater than or equal to 13.

| Justify that 2011 is a prime number. It follows that card $B=1$ | $\mathbf{1 p}$ |
| :--- | :--- |


| b) $\frac{a(a+1)}{2}+\frac{b(b+1)}{2}=1+\frac{2}{c+1}$ | $\mathbf{1 p}$ |
| :--- | :--- |


| $a(a+1)$ and $b(b+1)$ are products of consecutive numbers $\Rightarrow \frac{2}{c+1} \in \mathbb{N}$, hence $c \in\{0 ; 1\}$ | $\mathbf{1 p}$ |
| :--- | :--- |

Discussion and finalization.

2p