4. How to form a rectangle, with all sides greater than 1, from 13 rectangles of sizes $1 \times 1, 2 \times 1, 3 \times 1, \ldots, 13 \times 1$?
Answer. One of the possible examples is shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_06_ddc950af38f62db5075bg-2.jpg?height=394&width=708&top_left_y=771&top_left_x=137)

Comment 1 (how to guess the example). If we group the rectangles: the first with the last, the second with the second-to-last, and so on, we get identical strips of length 14: $1+13=14, 2+12=14, 3+11=14, \ldots$ It is easy to form a rectangle from several identical strips by attaching them to each other along the long side. However, if we try to continue the process described above, a problem arises: the central strip $7 \times 1$ will remain without a pair. This problem can be solved as follows: leave the longest strip of length 13 alone, and group the remaining strips in the same way (the shortest with the longest, etc.): 13, 1+12, 2+11, 3+10, 4+9, 5+8, 6+7. By attaching the resulting strips of length 13 to each other, we get the example shown above.

The same idea was used by the young Gauss to find the sum of numbers from 1 to 100.

Comment 2. The area of the final rectangle is $1+2+3+4+\ldots+13=91$. Since 91 has only two prime factors in its factorization: 91=7$\cdot$13, the sides of the final rectangle are uniquely determined - they must be 7 and 13. It is impossible to form a rectangle of different dimensions (such that its sides are greater than 1) from the given rectangles. However, the initial rectangles can be arranged differently within the final rectangle. Some variants of the arrangement are shown below:
![](https://cdn.mathpix.com/cropped/2024_05_06_ddc950af38f62db5075bg-2.jpg?height=706&width=1580&top_left_y=2050&top_left_x=136)

All-Russian School Olympiad 2014-2015

## Grading Criteria.

- Correct drawing (no justifications or reasoning required) - 7 points.
- Complete description of how to form the desired rectangle (for example, such as this: from rectangles $1 \times 1$ and $12 \times 1$ form a rectangle $13 \times 1$. Form the same rectangles from $2 \times 1$ and $11 \times 1, 3 \times 1$ and $10 \times 1, \ldots, 6 \times 1$ and $7 \times 1$. Then, from seven rectangles $13 \times 1$ form a rectangle $13 \times 7$) - 7 points.
- The idea of grouping strips as "first-last, second-second-to-last," not fully developed - 1-2 points.