Elementary Mathematics Dorofeev |best| May 2026

Now remove the top-left corner (1,1). Its color is (1+1) mod 3 = 2 mod 3 = Color 2? Wait — careful: (1+1)=2, so 2 mod 3 = 2 — yes, Color 2. So after removal: Color 0: 9 Color 1: 8 Color 2: 7 (since we removed one from Color 2) Each 1×3 tromino, no matter how you place it (horizontal or vertical), covers exactly one square of each color .

We have (9 instead of 8) and too few Color 2 (7 instead of 8). Impossible. 6. The Beautiful Conclusion The tiling fails not because of a bad arrangement, but because of an invariant — a numerical property preserved by every tromino but violated by the board’s initial coloring counts. elementary mathematics dorofeev

Thus, . 5. The Contradiction If 8 trominoes tile the shape, they would cover: 8 trominoes × 1 square of each color = 8 of Color 0, 8 of Color 1, 8 of Color 2. Now remove the top-left corner (1,1)

Try to visualize: the 5×5 board has 25 squares. Remove one corner → 24 squares. Each tromino covers 3 squares. 24 ÷ 3 = 8 trominoes needed. So numerically it’s possible. So after removal: Color 0: 9 Color 1:

But ? 2. The First Attempt You try. Place a tromino horizontally in the top row. Then another. You quickly get stuck — the missing corner leaves an awkward gap. After some attempts, you suspect it’s impossible .