(a) A move changes $S$ by $0 \pmod{3}$, but the final sum is always a multiple of $6$ i.e. $\equiv 0\pmod{3}$, which means $S$ has to be a multiple of $3$, contradicting the "independent of S" condition. (b)Lemma: All $2^n \times 2^n$ boards with a square removed from a corner can be tiled by $L$-triominos. Proof: Easy induction We now prove the problem for all powers of $2$. Base case is $n=2$. Let the four numbers be $a>b>c>d$. Realize that adding $1$ to three of numbers is like subtracting one from the remaining number, so we obviously can subtract until all terms are equal. For the inductive step, we split the $2^n \times 2^n$ into four $2^{n-1}\times 2^{n-1}$. By the hypothesis, we can make all elements equal in each quadrant. We then make all the terms equal in the middle $2\times 2$ square. Realize that we have effectively taken away a square from each quadrant. Each quadrant has all terms equal except that one corner. We may now apply our $L-$triomino tessellation from the lemma to each quadrant and add 1's to each triomino until they are all equal to the number in their respective corners, making the whole board equal.