Compare the parity of the sum of all elements before and after one operation.
If the chosen numbers are a,b, the operation removes them and appends β£aβbβ£. In parity,
β£aβbβ£β‘aβbβ‘a+b(mod2).
Therefore, the parity of the sum of all elements never changes.
The initial sequence is (1,2,β―,2x), so its sum is
1+2+β―+2x=x(2x+1).
Since 2x+1 is odd,
x(2x+1)β‘x(mod2).
Thus, the parity of the last remaining element is equal to the parity of x.
So f(x)β‘x(mod2). From Piββ‘riβ(mod2), we get
Kiβ=1βPiβ.
After converting the bit string K to an integer, compute
N=CβK.
Solution written by GPT5.5