Invariants I

The initial number is even, and adding \(2\) preserves parity. The number \(99\) is odd, so it cannot be obtained.

The number of heads changes by \(-2\), \(0\), or \(2\), so its parity is preserved. It starts at \(9\) and would need to become \(0\). Odd cannot become even.

Initially there are \(15\) stones, while the target has \(20\). Since moving a stone does not change the total, the target is impossible.

The residue modulo \(3\) is preserved. \(5\equiv2\pmod3\), while \(100\equiv1\pmod3\). Therefore \(100\) cannot be obtained.

Changing two signs changes the number of minuses by \(-2\), \(0\), or \(2\). Its parity is preserved. Initially it is \(0\), but the target is \(3\), impossible.

Changing two signs multiplies the product by \((-1)^2=1\). Initially the product is \(+1\). With \(7\) minuses, the product is \(-1\). Impossible.

Initially there are \(14\) tokens, while the target has \(15\). Moving a token preserves the total, so the target is impossible.

If \(h\) heads are among the flipped coins, the number of heads changes by \(4-2h\), an even number. Parity is preserved. Initially \(15\) is odd; \(2\) is even. Impossible.

The initial sum is \(11\), the target sum is \(15\). The operation preserves the sum, so it is impossible.

In chessboard coloring, diagonally adjacent squares have the same color. Thus the token's color is preserved. It cannot move from black to white.

Changing \(4\) signs multiplies the product by \((-1)^4=1\). Initially the product is \(+1\), while with one minus it is \(-1\). Impossible.

In one move, the sum changes by \(1+1-2=0\). Thus the sum itself is preserved, and therefore its residue modulo \(3\) is preserved.

If \(h\) of the chosen lamps are on, the number of on lamps changes by \(100-2h\), an even number. Initially \(0\) lamps are on, even; finally \(2025\) would be on, odd. Impossible.

In one move, the total number of tokens changes by \(4-1=3\), so its residue modulo \(3\) does not change.

The original board has \(32\) black and \(32\) white cells. Opposite corners have the same color, so after removal there are \(30\) cells of one color and \(32\) of the other. Each domino covers one black and one white cell, so tiling is impossible.

Each move changes \(4\) signs, so the product of all \(16\) signs is multiplied by \((-1)^4=1\). Initially the product is \(+1\). With exactly one minus, it is \(-1\). Impossible.

The sum \(1+2+\cdots+10=55\) is odd. Changing the sign of \(a\) changes the sum by \(-2a\), an even number. Thus the parity of the sum remains odd. Zero is even, so it is impossible.

A knight move changes the square color. After an odd number of moves, the color is opposite to the initial one. Since \(2025\) is odd, the knight is on a black square, not white.

Parity is preserved because even numbers are added. The initial number \(1\) is odd, while \(100\) is even. Impossible.

Each move increases \(x+y\) by \(3\). Thus \(x+y\) modulo \(3\) is preserved and equals \(0\). At \((10,10)\), the coordinate sum is \(20\equiv2\pmod3\). It cannot be reached.

The product of all \(36\) entries is preserved: one move multiplies it by \((-1)^6=1\). Initially the product is \(1\). A table with one \(-1\) has product \(-1\). Thus it is unreachable.

Every knight move changes the square color. After \(15\) moves, since \(15\) is odd, the knight is on the opposite color. The starting square is black, so it cannot return to it.

One move changes the sum by \(2+2-1=3\), so the sum modulo \(3\) is preserved. Initially the sum is \(3\equiv0\pmod3\), finally \(30\equiv0\pmod3\), so this invariant does not forbid it. In fact the state is reachable: apply three moves, subtracting from a different coordinate each time; all coordinates increase by \(3\). Repeating this three-move cycle three times gives \((10,10,10)\).

The initial number of inversions is \(0\). Each adjacent swap changes inversion parity. After \(27\) moves, inversion parity would be odd. But in \(87654321\), the number of inversions is \(C(8,2)=28\), which is even. Contradiction; exactly \(27\) moves are impossible.