7.4. Anton from the village was given several zucchini, and he decided to give them to his friends. He gave half of the received zucchini to Arina, and a third (also from the received amount) to Vera. It turned out that after this, the number of zucchini Arina had became a square of some natural number, and the number of zucchini Vera had became a cube (they had no zucchini before). Find the smallest possible number of zucchini that Anton could have received from the village. (Find the answer and prove that it is minimal).
Solution. Let Anton receive $n$ zucchinis. Since both half and a third of $n$ are integers, $n$ is divisible by 6, that is, $n=6k$ for some natural number $k$. Then it is known that $3k$ is the square of a natural number, and $2k$ is a cube. Let $k=2^p 3^q m$. In other words, let $p$ and $q$ be the highest powers of two and three, respectively, by which $k$ is divisible. Then $3k=2^p 3^{q+1} m$, and since this is a square, $p$ and $q+1$ are even numbers. On the other hand, $2k=2^{p+1} 3^q m$ is a cube, so $p+1$ and $q$ are divisible by 3. Note that then $p$ is at least 2 (because $p+1$ is divisible by 3), and $q$ is at least 3 (because $q$ is divisible by 3 and $q \neq 0$ from the condition of the evenness of $q+1$). Then $k$ is divisible by at least 4 and 27, so $k$ is at least 108, and $n$ is at least 648. It is clear that $648=6 \times 4 \times 27$ fits the construction.

Criteria. Only the answer - 1 point.

Verification that the answer fits - 1 point (in this solution, this verification implicitly follows from the fact that we are looking for exactly the suitable numbers).

Proven that $n$ is divisible by 6 - 1 point.

Proof that $k$ must be divisible by 4 and 27 (or that $n$ is divisible by 8 and 81, which is the same) - 2 points for each fact.

All the above points are cumulative.

The answer can be left in the form of a product; no points are deducted for the lack of an explicit notation.