2. a) Determine the natural numbers of the form $\overline{a b a}$, written in the decimal system, with the property:

$$
\sqrt{\overline{a b a}}=\frac{a^{2}}{3}+2 b
$$

b) Determine the non-zero natural number $x$ with the property:

$$
\sqrt{(1 \cdot 2 \cdot 3 \cdot \ldots \cdot x)^{2}+13} \in \mathbb{N}
$$
Solution: a) Since $a$ and $b$ are digits, $a \neq 0$, we have $\frac{a^{2}}{3}+2 b \in \mathbb{Q} \Rightarrow \sqrt{abc} \in \mathbb{Q} \Rightarrow \overline{abc}=n^{2}, n \in \mathbb{N}$. Therefore, $\frac{a^{2}}{3}+2 b \in \mathbb{N} \Rightarrow 3 \mid a^{2}$, and since 3 is a prime number, it follows that $a=3k, k \in \mathbb{N}$. We have: $a \in\{3,6,9\}$. For $a=3$, we have $\sqrt{3b3}=2b+3$, and the only perfect squares between 300 and 400 are $18^{2}=324$ and $19^{2}=361$, which do not satisfy the condition. For $a=6$, we have $\sqrt{6b6}=2b+12$, and the perfect squares between 600 and 700 are $25^{2}=625$ and $26^{2}=676$. It is verified that $\overline{aba}=676$. For $a=9$, we have $\sqrt{9b9}=2b+27$, and the perfect square between 900 and 999 is $31^{2}=961$, which does not satisfy the condition.

b) $\sqrt{(1 \cdot 2 \cdot 3 \cdot \ldots \cdot x)^{2}+13} \in \mathbb{N} \Rightarrow(1 \cdot 2 \cdot 3 \cdot \ldots \cdot x)^{2}+13$ is a perfect square. Since the remainder of a perfect square when divided by 5 is 0, 1, or 4, for $x \geq 5$, the number $(1 \cdot 2 \cdot 3 \cdot \ldots \cdot x)^{2}+13$ is not a perfect square because its remainder when divided by 5 is 3. The cases to be analyzed are $x \in\{1,2,3,4\}$. The only case that satisfies is $x=3$.

## Grading:

| a) $a$ and $b$ are digits, $a \neq 0$, we have $\frac{a^{2}}{3}+2 b \in \mathbb{Q} \Rightarrow \sqrt{\overline{abc}} \in \mathbb{Q} \Rightarrow \overline{abc}=n^{2}, n \in \mathbb{N}$  $\frac{a^{2}}{3}+2 b \in \mathbb{N} \Rightarrow 3 \mid a^{2}$, and since 3 is a prime number, it follows that $a=3k, k \in \mathbb{N}$. We have: $a \in\{3,6,9\}$ | $1 \text{p}$ |
| :---: | :---: |
| For $a=3$ we have $\sqrt{\overline{3b3}}=2b+3$, and the only perfect squares between 300 and 400 are  $18^{2}=324$ and $19^{2}=361$ which do not satisfy the condition | $1 \text{p}$ |
| For $a=6$, we have $\sqrt{6b6}=2b+12$, and the perfect squares between 600 and 700 are $25^{2}=625$  and $26^{2}=676$. It is verified that $\overline{aba}=676$. | $1 \text{p}$ |
| For $a=9$, we have $\sqrt{\overline{9b9}}=2b+27$, and the perfect square between 900 and 999 is $31^{2}=961$ which  does not satisfy the condition. | $1 \text{p}$ |
| b) the remainder of a perfect square when divided by 5 is 0, 1, or 4 | 1 p |
| for $x \geq 5$, the number $(1 \cdot 2 \cdot 3 \cdot \ldots \cdot x)^{2}+13$ is not a perfect square because its remainder when divided by 5 is 3 | $1 \text{p}$ |
| The cases to be analyzed are $x \in\{1,2,3,4\}$. The only case that satisfies is $x=3$. | $1 \text{p}$ |