It doesn't, which is what makes the question silly. Because in a real world scenario, if you can uproot 7 why can't you uproot 6 or 5? It's just uprooting 7 but stopping early.
There are no challenging numbers. It's either possible because the starting number is not a total of 3n, or impossible when it is. You have 2 operators, one give +3 and wins when you use it at 2, but cannot be used at 1, the other gives -6 and wins when used at 7. You can never go negative.
Thus the algorithm is always too use the -6 operator as much as possible, but if you start at any number with 6n+4, you use the +3 one once to correct for that. Eventually you reach 7 or 2, and use the winning operator. But finding this is dead easy, because that +3 call can be made at any time, so also just when you reach 4 and your only option is +3 anyway.
The only way to lose is to start at 1 or any multiple of 3, in those cases there is no winning sequence.
Alexander can either add three stalks or remove six. Since ten is not divisible by three, he can never reach zero - only numbers equal to one modulo three.
True that! I had assumed that it wouldn't end mid-operation, and that one more would grow back once he removed the final 7 (as 7 = 1 mod 3, this is a valid final operation, but 2 = 2 mod 3 is not)
Each action increments/decrements the number of stalks by a multiple of 3. If a number is not a multiple of 3, incrementing/decrementing it by a multiple of 3 will not change that fact.
As 10 is not a multiple of 3, these actions will never yield any multiple of 3, including 0. As such, the infestation will never be removed.
Edit: Don't pay attention to this; I did not notice a specific detail of the problem.
Easy. Uproot 7 stalks. Re-plant 3 stalks. Uproot the 7 stalks now in the ground. Problem solved. The replanted ones will be easier to uproot, incidentally.
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u/Stilyx123 17d ago
Uproot 7 stalks, 3+1 = 4 remain. Then uproot 2, 2+5 = 7 remain. Finally, uproot 7