Problem 5 · 1991 AJHSME
Medium
Geometry & Measurement
paritytiling
A “domino” is made up of two small squares:
Which of the “checkerboards” illustrated below CANNOT be covered exactly and completely by a whole number of non-overlapping dominoes?
(A)
(B)
(C)
(D)
(E)
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Answer: B — 3 × 5.
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Hint 1 of 3
You don't need to attempt any actual covering. One domino always hides exactly 2 squares — so the squares always vanish two at a time. What must be true about the TOTAL number of squares for them all to disappear in pairs?
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Hint 2 of 3
Dominoes cover squares in twos, so a fillable board must have an EVEN count of squares. The answer is the lone board whose square-count is odd — and square-count is just rows × columns.
Still stuck? Show hint 3 →
Hint 3 of 3
Multiply rows × columns for each board and hunt for the odd one — odd × odd is the only way to get an odd total.
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Approach: a board fillable by dominoes must have an even number of squares
- Each domino covers exactly 2 squares, so however you lay them, squares get used up two at a time. That means a board can be filled only if its total number of squares is even — no clever arrangement can save an odd-square board.
- Count the squares (rows × columns): 3×4 = 12, 3×5 = 15, 4×4 = 16, 4×5 = 20, 6×3 = 18. Only 15 is odd.
- An odd number of squares can never split into pairs, so 3 × 5 is the board that cannot be covered.
- Why this transfers: any tile that covers a fixed number of cells forces the total to be a multiple of that number — a divisibility check often settles "can it be tiled?" instantly, before you ever try to fit a single piece.
- Going deeper (the harder version): an even count is necessary but not always enough. Color the board like a checkerboard — each domino must cover one dark and one light square, so a board with unequal dark/light counts is impossible even when its total is even. Here, though, the simple even/odd count already names the answer.
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