4. Determine the natural numbers $a, \overline{bcd}$ and $e$, knowing that $4 \cdot\left(2^{2 a} \cdot 3^{a}+\overline{bcd}\right)+2^{e}=2013$.

## SOLUTIONS

## 5th Grade
4. Since $4 \cdot\left(2^{2 a} \cdot 3^{a}+\overline{b c d}\right)$ is even and 2013 is odd, it follows that $2^{e}$ is odd, so $e=0$. Then $2^{2 a} \cdot 3^{a}+\overline{b c d}=503$, that is, $12^{a}+\overline{b c d}=503$. We deduce that $a \leq 2$ and, considering the three cases, we find the solutions $(a, \overline{b c d}) \in\{(0,502),(1,491),(2,359)\}$.