An invisible tank is on a $100 \times 100 $ table. A cannon can fire at any $k$ cells of the board after that the tank will move to one of the adjacent cells (by side). Then the progress is repeated. Find the smallest value of $k$ such that the cannon can definitely shoot the tank after some time.
Problem
Source: 239 2017 S7
Tags: combinatorics
22.05.2020 14:14
What is the size of the tank? $1 \times 1$ ? Also, is the answer $51$? @2below Is time a constraint?
22.05.2020 15:27
Is their some condition like the tank cannot trace a cell twice?
22.05.2020 15:32
Isn't the answer 1? Assume that the cannon takes 5 minutes to fire at any one cell. Then it takes 10 minutes for the tank to move to an adjacent square. You are firing at 96 squares per day. Theoretically, the cannon should hit the tank after some time. I'm probably wrong though.
22.05.2020 16:37
i have got a upper bound of 15 so far by construction. And it's definitely not 1(any box has atleast 2 ways out)
23.05.2020 10:38
Dem0nfang wrote: What is the size of the tank? $1 \times 1$ ? Obviously. Quote: @2below Is time a constraint? Not at all. The tank moves immediately after a shot. A-Thought-Of-God wrote: Is their some condition like the tank cannot trace a cell twice? No.
21.05.2021 06:51
This problem appears as theorem 3 in the following paper: http://www.math.bas.bg/moiuser/~ACCT2016/a7.pdf
21.05.2021 19:31
After a tank shoots at $k$ spots, will it never shoot those $k$ spots again? Also, does the tank start anywhere? Or at the first cell?
22.08.2021 18:28
The problem is Intersting....... Bummp
30.08.2021 17:52
One can use St Petersburg MO 2017 9.7. Just see on every turn how many possible cells are there for the tank before it moved (this works even if we suppose we know the color of the cell of the tank in the beginning (supposing we have a checkerboard coloring)). This is for the proof that $k=50$ does not work.
18.07.2023 11:11
can anyone provide construction for k=51?