Note that for $n=2000=40 \cdot 49+40$ the required cutting exists (left figure).
![](https://cdn.mathpix.com/cropped/2024_05_06_7e3e695203bdf41f05f0g-27.jpg?height=296&width=690&top_left_y=2464&top_left_x=682)

Assume that there exists a square $n \times n$, where $n>2000$. Contradiction.

## Answer

$n=2000$
Zamyatin $B$.

For what least $n$ can a square $n \times n$ be cut into squares $40 \times 40$ and $49 \times 49$ so that squares of both types are present?