Solution: The total area of the hall is $12 \times 12 = 144$ square meters. Each rectangle has area at least 18, so the maximum number of rectangles is $ \left\lfloor \frac{144}{18} \right\rfloor = 8 $, but we want the *minimum* number, so we seek the largest rectangle area that divides 144 and is $ \geq 18 $, and check if tiling with such rectangles covers the square.

["Solution: Tiling a 12×12 Hall with Rectangles of Area at Least 18", "When tasked with tiling a total area of $144$ square meters—such as a square hall measuring $12 \ imes 12$ meters—using rectangular tiles with an area of at least $18$ square meters, the key challenge is determining the minimum number of rectangles needed to completely cover the space. This involves both number theory (finding valid rectangle dimensions) and geometric tiling principles.", "### Step 1: Determine the Maximum Rectangle Area Possible", "Given the constraint that each rectangle must have area at least $18$, the theoretical upper bound for the number of tiles is:", "$$\n\left\lfloor \frac{144}{18} \right\rfloor = 8\n$$", "However, achieving this bound requires tiling the entire $12 \ imes 12$ hall using eight rectangles, each with area exactly $18$. Can such a tiling be constructed?", "### Step 2: Find Rectangle Dimensions with Area ≥ 18", "We seek rectangle dimensions $a \ imes b$ such that $a \cdot b \geq 18$, where both $a$ and $b$ divide $12$ (or fit evenly into $12$ to allow tiling without gaps), to maximize area efficiency:", "Possible rectangle pairs $(a,b)$ with width $a, b \mid 12$:", "- $1 \ imes 12 = 12$ → too small\n- $2 \ imes 6 = 12$ → too small\n- $3 \ imes 6 = 18$ → valid\n- $4 \ imes 4 = 16$ → too small\n- $3 \ imes 4 = 12$ → too small\n- $3 \ imes 8 = 24$ → too big (exceeds 144 total area if used)\n- $4 \ imes 5 = 20$, but $5 <br/>\nmid 12$ → hard to tile\n- $6 \ imes 6 = 36$, area $36 \geq 18$ → valid", "Indeed, $3 \ imes 6 = 18$ is the largest rectangle area (in area) that divides $144$ evenly and fits geometrically in a $12 \ imes 12$ square when properly arranged.", "### Step 3: Attempt Tiling with $3 \ imes 6$ Rectangles", "Area per tile: $3 \ imes 6 = 18$\nMax number of tiles: $144 / 18 = 8$", "Can we tile a $12 \ imes 12$ square with eight $3 \ imes 6$ rectangles?", "Yes — consider partitioning the hall into two $6 \ imes 12$ halves:", "- The full $12 \ imes 12$ square fits two $6 \ imes 12$ strips.\n- Each $6 \ imes 12$ band can be divided into four rectangles of $3 \ imes 6$:\n - First $3$ meters wide: place four $3 \ imes 6$ tiles side by side.\n - Next $3$ meters wide: repeat.", "Thus, $2 \ ext{ strips} \ imes 4 = 8$ rectangles total, exactly covering the hall.", "### Step 4: Is 8 Truly the Minimum?", "Could fewer than 8 rectangles suffice, each with area $\geq 18$?", "The theoretical minimum number is:", "$$\n\left\lceil \frac{144}{144} \right\rceil = 1 \ ext{ (if one tile of } 144\ ext{ exists)}\n$$", "But $144$ rectangle of size $1 \ imes 144$ or $12 \ imes 12$ cannot form a valid tiling across a square layout without extreme irregularity, and certainly not dividing evenly into physically arranged rectangles within the $12 \ imes 12$ constraint.", "Any rectangle larger than $18$ (e.g., $24$, $36$, $48$) exceeding 144 when used once invalidates the total area. The next possible area is $24$ ($4 \ imes 6$), but $144 / 24 = 6$. So 6 tiles of area 24 are possible?", "Area $24$ rectangles: $4 \ imes 6 = 24$, $144 / 24 = 6$", "Can a $12 \ imes 12$ square be tiled with six $4 \ imes 6$ rectangles?", "Each $4 \ imes 6$ tile has area $24 > 18$, satisfying the constraint.", "Try arrangement:\n- Divide the $12 \ imes 12$ square into three $4 \ imes 12$ vertical strips.\n- Each strip is $12 \ imes 4$.\n- A $4 \ imes 6$ tile requires $4 \ imes 6$, but $6$ does not divide $12$, so we cannot perfectly tile a $4 \ imes 12$ strip: $12 / 6 = 2$, so only two $4 \ imes 6$ tiles side-by-side, leaving $4 \ imes 6$? No — actually, $12 \div 4 = 3$ along length, and $6$ fits twice in 12, so $3 \ imes 2 = 6$ tiles per strip. Each strip holds $12$ tiles? Wait — no: each tile is $4 \ imes 6$, so placed with 6m side across 12m length: fits 2 tiles (6×2=12), each row of 2 tiles. Each strip is $12 \ imes 4$, assigning height $4$, so 3 such strips vertically: $3 \ imes 2 = 6$ tiles total per strip? No — per strip (height 4), we place tiles with 6m along width: $12 / 6 = 2$ tiles per row, and 3 rows (height 4), so $2 \ imes 3 = 6$ tiles per strip. With 3 strips: $3 \ imes 6 = 18$ — too many.", "Wait — mistake: we want only one such tile per rectangle? No, we already know 6 rectangles of area 24: $144 / 24 = 6$, so total area matches.", "But can 6 rectangles of size $4 \ imes 6$ fully tile $12 \ imes 12$?", "Try dividing the square into three horizontal strips of $12 \ imes 4$. Each $12 \ imes 4$ section:", "- Area: $48$ — perfect for two $4 \ imes 6 = 24$ tiles? $48 / 24 = 2$ — yes.", "But $4 \ imes 6$ won’t fit in $12 \ imes 4$ vertically if oriented vertical: height 6 > 4 → no.", "Rotate: place tiles as $6 \ imes 4$. Then in $12 \ imes 4$:", "- Along 12m width: $12 / 6 = 2$ tiles\n- Along height: $4 / 4 = 1$ row\n→ $2 \ imes 1 = 2$ tiles per such rectangle.", "To get 6 total tiles: $6 / 2 = 3$ such rectangles → only cover $12 \ imes 12$ area $12 \ imes 12 = 144$? No: 3 × 2 = 6 tiles × 24 = 144 — correct.", "But 3 rectangles × (size $6 \ imes 4$) = $3 \ imes 24 = 72$, insufficient. Wait — inconsistency.", "Actually, $6 \ imes 24 = 144$, so need 6 rectangles of area 24.", "Each $6 \ imes 4 = 24$ → fits in $12 \ imes 12$:", "Divide into three $12 \ imes 4$ horizontal strips. In each, place two $6 \ imes 4$ tiles side by side (6+6=12). Each strip holds 2 tiles → total $3 \ imes 2 = 6$ tiles. Perfect.", "Thus, 6 rectangles of size $6 \ imes 4$ (area 24) tile the hall—each area $> 18$, and total count is minimized at 6.", "But earlier we found 8 with $3 \ imes 6$. Is 6 valid?", "Yes. So is 8 not minimal? But wait — can we do better than 6?", "Try $9 \ imes \frac{11.11}$? Not integer.", "Largest rectangle area under 144 with integer sides dividing 12 and area ≥ 18:", "- $3 \ imes 6 = 18$ → 8 tiles → area 144\n- $4 \ imes 6 = 24$ → 6 tiles\n- $3 \ imes 8 = 24$, but 8 doesn’t divide 12 vertically\n- $4 \ imes 9 = 36$ → 4 tiles\n- $6 \ imes 6 = 36$ → 4 tiles\n- $9 \ imes 16$? Too big, impractical\n- $12 \ imes 12$: 1 tile, but area 144 ≥ 18, but is it a rectangle? Yes — but number is 1, which is less than 6.", "Wait — is a single $12 \ imes 12$ tile allowed?", "Yes! Area is $144 \geq 18$, and it perfectly covers the hall.", "So number of rectangles = 1.", "But the problem says: “each rectangle has area at least 18” — $144 \geq 18$, so valid.", "Then minimum number is 1, achieved by the hall itself as a single rectangle.", "But is this meaningful? The problem says “rectangles” — a square is a rectangle — so allowed.", "But likely, the intent is to partition into multiple rectangles (greater than one), but the problem does not specify.", "Re-reading: “the maximum number of rectangles is… but we seek the minimum number…”", "So minimum possible.", "Then: one rectangle of size $12 \ imes 12$ satisfies all constraints: area = 144 ≥ 18, fits exactly.", "But is this considered a valid tiling? Yes.", "However, the earlier step suggested using multiple smaller rectangles — but the theoretical upper bound on count is 8, but minimum is 1.", "But that"]