Bogdanov I.I.

For some 2011 natural numbers, all their 2011$\cdot$1005 pairwise sums were written on the board.

Could it be that exactly one third of the written sums are divisible by 3, and another exactly one third of them give a remainder of 1 when divided by 3?
Let's consider, for example, the numbers from 1 to 2011. Among them, $2010: 3=670$ numbers are divisible by 3, the same number give a remainder of 2 when divided by 3, and 671 numbers give a remainder of 1. Therefore, $670 \cdot 669: 2 + 670 \cdot 671 = 335 \cdot 2011 = 2011 \cdot 1005$: 3 sums are divisible by three and the same number give a remainder of 1.

## Answer

It could.

Author: $\underline{\text { Folklore }}$

A rope was folded in half, then again in half, and then once more in half, and then all layers of the rope were cut in one place.

What could have been the length of the rope if it is known that some two of the resulting pieces had lengths of 9 meters and 4 meters?

## Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_b3d98ec29b07c8345931g-35.jpg?height=258&width=810&top_left_y=500&top_left_x=633)

Upon cutting, pieces of three different types could have been formed (see the figure), with the length of one type of piece exactly twice that of another. Let's denote these lengths as $x, 2x$, and $y$ respectively, and the length of the entire rope as $L$. By counting the number of pieces of each type, we get that

$L = 8x + 4y = 4(2x + y)$. All possible lengths of the pieces, based on the condition, are presented in the table:

| $x$ | 4 | 2 | 9 | 4.5 |
| :---: | :---: | :---: | :---: | :---: |
| $y$ | 9 | 9 | 4 | 4 |
| $L$ | 68 | 52 | 88 | 52 |

## Answer

52 meters, 68 meters, or 88 meters.