A neighborhood consists of $10 \times 10$ squares. On New Year's Eve it snowed for the first time and since then exactly $10$ cm of snow fell on each square every night (and snow fell only at night). Every morning, the janitor selects one row or column and shovels all the snow from there onto one of the adjacent rows or columns (from each cell to the adjacent side). For example, he can select the seventh column and from each of its cells shovel all the snow into the cell of the left of it. You cannot shovel snow outside the neighborhood. On the evening of the 100th day of the year, an inspector will come to the city and find the cell with the snowdrift of maximal height. The goal of the janitor is to ensure that this height is minimal. What height of snowdrift will the inspector find? Proposed by A. Solynin
Problem
Source: All-Russian MO 2024 9.5=11.5
Tags: combinatorics, combinatorics proposed
07.06.2024 21:19
Answer: $1120 \text{ cm}$ Bound: Notice on the final day $10$ squares will have no snow in them. So there must exist a square with at least $(10\cdot 100\cdot 100)/(100-10)>1111$ centimeters of snow (we are done as the amount of snow in each cell is a multiple of $10$). Construction: We will just be moving rows. Let one row have $10a$ centimeters in each cell and an adjacent row have $10b$ centimeters in each cell. We define cycling these two rows as shoveling back and forth between these rows for $k$ days resulting in $10a+10b+20k$ in all the cells of one row and $0$ in the other. We can cycle the first two rows so that the first ends up with a certain amount of snow in each cell such that if left untouched will either be $1110$ or $1120$ centimeters. We can then continue on cycling between the second and third and so on (clearly the next pair of rows doesn't already have enough snow that if untouched would result in $1110$ centimeters).
11.06.2024 05:34
sami1618 wrote: Answer: $1120 \text{ cm}$ Bound: Notice on the final day $10$ squares will have no snow in them. So there must exist a square with at least $(10\cdot 100\cdot 100)/(100-10)>1111$ centimeters of snow (we are done as the amount of snow in each cell is a multiple of $10$). Construction: We will just be moving rows. Let one row have $10a$ centimeters in each cell and an adjacent row have $10b$ centimeters in each cell. We define cycling these two rows as shoveling back and forth between these rows for $k$ days resulting in $10a+10b+20k$ in all the cells of one row and $0$ in the other. We can cycle the first two rows so that the first ends up with a certain amount of snow in each cell such that if left untouched will either be $1110$ or $1120$ centimeters. We can then continue on cycling between the second and third and so on (clearly the next pair of rows doesn't already have enough snow that if untouched would result in $1110$ centimeters). Wow that's a clever solution