AMC 8 · 2024 · #7

• First split the problem by counting squares.
• The board has $3\times 7 = 21$ unit squares.
• Each $2\times 2$ tile and each $1\times 4$ tile covers exactly 4 squares, so the 'big tiles' together cover a multiple of 4 squares.
• Whatever is left over must be covered one square at a time by $1\times 1$ tiles.
• So the number $c$ of $1\times 1$ tiles equals 21 minus a multiple of 4.

💡 Grade 4 area formula $\text{length}\times\text{width}$ gives 21 squares, and the problem reduces to one short sentence: 'big-tile squares + small-tile squares $= 21$.'

• Now divide 21 by 4.
• We have $21 = 4\times 5 + 1$, so the remainder is 1.
• That means $c$ must leave remainder 1 when divided by 4: $c = 1$ when 5 big tiles are used, $c = 5$ when 4 big tiles are used, $c = 9$ when 3 big tiles are used, and so on.
• Among the answer choices (A)–(E), only 1 and 5 are remainders-of-1 mod 4.
• So the answer is either (A) 1 or (E) 5.

💡 Sorting numbers by their remainder when divided by 4 is exactly the Grade 4 idea of multiples — no algebra needed.

• Draw the $3\times 7$ grid (Tool #1) and try the smaller candidate $c=1$ first using Tool #6 (guess & check).
• With $c=1$, we need 5 big tiles to cover the other 20 squares.
• Because the grid is only 3 squares tall, every $1\times 4$ tile sits horizontally inside a single row, and every $2\times 2$ tile sits either in the top-two rows or the bottom-two rows.
• No matter how we slide these 5 big tiles onto the diagram, they always leave more than one straggling square — typically a $2\times 1$ strip plus more — instead of exactly one hole.

💡 Grade 4 students classify how rectangles, parallel rows, and perpendicular columns fit together — exactly the diagram check needed to see why a single hole can't survive.

• To pin down WHY $c=1$ fails, color the columns of the diagram alternately Red, Blue, Red, Blue, $\ldots$, starting Red.
• With 7 columns, 4 are Red and 3 are Blue.
• Each column has 3 squares, so the diagram has $4\times 3 = 12$ red squares and $3\times 3 = 9$ blue squares — Red exceeds Blue by exactly 3.
• Now look at the big tiles on this picture: a $2\times 2$ tile sits in two adjacent columns (one red, one blue), covering 2 reds + 2 blues; a $1\times 4$ tile spans 4 consecutive columns (2 reds + 2 blues), also covering 2 reds + 2 blues.
• Every big tile covers equal red and blue squares, so big tiles can never close the Red-minus-Blue gap of 3.
• That gap must be closed by $1\times 1$ tiles: at least 3 more $1\times 1$ tiles must land on red squares than on blue squares.
• With only $c=1$ total, that is impossible.
• So $c=1$ cannot work.

💡 Coloring the diagram turns the impossibility argument into a Grade 4 multi-step word problem about how many more red squares than blue squares are left over.

• Since $c=1$ is impossible, try the next candidate $c=5$ with Tool #6.
• On the diagram, place three $2\times 2$ tiles (A, C, D) covering the top two rows of columns 1–6, then one $1\times 4$ tile (B) covering the bottom row of columns 1–4.
• The remaining 5 squares — the entire column 7 (3 squares) and the bottom of columns 5–6 (2 squares) — are filled with five $1\times 1$ tiles.
• The board is fully covered with $c=5$.

💡 Drawing the tile arrangement directly on the diagram confirms in one glance that $c=5$ really is achievable.

• Putting it together: the allowed values of $c$ are $1, 5, 9, \dots$; the value $c=1$ is ruled out by the red-blue gap; and $c=5$ is realized by the explicit diagram above.
• So the minimum number of $1\times 1$ tiles is 5, giving answer (E).

💡 Two candidates, one impossible and one constructed — multi-step Grade 4 reasoning picks out the minimum.