Badzian A.I.

In a $10 \times 10$ square, numbers from 1 to 100 are arranged: in the first row - from 1 to 10 from left to right, in the second row - from 11 to 20 from left to right, and so on. Andrey plans to cut the square into dominoes $1 \times 2$, calculate the product of the numbers in each domino, and sum the resulting 50 numbers. He aims to achieve the smallest possible sum. How should he cut the square?
Let's number the dominoes in the partition. Let the numbers in the $i$-th domino be $a_{i}$ and $b_{i}$. Notice that $a_{i} b_{i}=\frac{a_{i}^{2}+b_{i}^{2}}{2}-\frac{\left(a_{i}-b_{i}\right)^{2}}{2}$. Summing these equalities over all dominoes, we get that the sum of all 50 products is $\frac{a_{1}^{2}+\ldots+a_{100}^{2}+b_{1}^{2}+\ldots+b_{100}^{2}}{2}-\frac{\left(a_{1}-b_{1}\right)^{2}+\ldots+\left(a_{100}-b_{100}\right)^{2}}{2}$. The first fraction is $1 / 2\left(1^{2}+\ldots+100^{2}\right)$, which does not depend on the partition. In the numerator of the second fraction, each square is either $1^{2}$ or $10^{2}$, depending on whether the $i$-th domino is horizontal or vertical. Therefore, the second fraction will be maximal (and the final sum minimal) when all terms in the numerator are equal to 100, that is, when all dominoes are vertical.

## Answer

On 50 vertical rectangles (see fig.).

![](https://cdn.mathpix.com/cropped/2024_05_06_388b3f8bbd0a17840215g-05.jpg?height=551&width=548&top_left_y=1714&top_left_x=755)

Send a comment