10. Kolya and Petya are dividing $2 n+1$ nuts, $n \geqslant 2$, with each wanting to get as many as possible. There are three ways of dividing (each goes through three stages).

1st stage: Petya divides all the nuts into two parts, each containing no fewer than two nuts.

2nd stage: Kolya divides each part again into two, each containing no fewer than one nut.

(1st and 2nd stages are common for all three ways.)

3rd stage: in the first way, Kolya takes the larger and the smaller parts; in the second way, Kolya takes both middle parts; in the third way, Kolya takes either the larger and the smaller parts or both middle parts, but in exchange for the right to choose, he gives Petya one nut.

Determine which way is the most advantageous for Kolya and which is the least advantageous for him.
10. Answer: the most profitable way for Kolya is the first one; with the second and third, he will get one nut less with correct play. (In general, as we will see, the dispute in this division is over one nut.)

No matter what piles (of $a$ and $b$ nuts, $a<b$) form after Petya's first move, Kolya can split the larger one into two parts of 1 and $b-1$ nuts, which will be the largest and the smallest, i.e., with the first division method, he can take $b \geqslant n+1$ nuts. (By taking $a=n, b=n+1$, Petya can prevent him from achieving more.) With the second division method, after the first move, $a=2, b=2n+1$, and with the best response $2=1+1$, $2n-1=n-1+n$, Kolya will get only $n$ nuts. (But with any other first move, he can get no less than $n+1$.) With the third method, Petya's move $a=n, b=n+1$ does not allow Kolya to achieve more than taking $n+1$ nuts (the sums of the two middle piles and the two outer ones will always be $n$ and $n+1$), so the extra nut that needs to be given away becomes decisive.