Sensim Math Original · sm-12
Easy mode Grade 3Problem
Picture a museum laid out as a 4×4 grid of rooms — 4 rows and 4 columns, so 16 rooms in all.
At night, two surveillance drones — Alpha and Beta — float in the air, each above exactly one room. While a drone is above a room, its cameras watch every room that is one step away horizontally, vertically, or diagonally. (So a drone above an inside room watches all 8 rooms around it. A drone above a corner room watches only 3 rooms.)
For tonight's schedule to be valid, two things must be true:
- The two drones must be above two different rooms.
- Neither drone is allowed to watch the room that the other drone is above.
Alpha and Beta carry different equipment, so swapping which drone is above which room counts as a different schedule.
How many valid (Alpha, Beta) schedules are there?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: On a $4 \times 4$ grid of $16$ museum rooms, place two distinguishable drones Alpha and Beta — one drone per room, two different rooms — so that neither drone is in a room watched by the other. "Watched" means horizontally, vertically, or diagonally one step away. Count the number of ordered (Alpha-room, Beta-room) schedules that satisfy this.
Givens: Grid: $4 \times 4 = 16$ display rooms; Each drone watches every room one step away (up to $8$ rooms, like a chess-king move); Alpha and Beta carry different sensors, so the assignment is ordered (Alpha at room $A$, Beta at room $B$ differs from the swap); The two drones occupy different rooms, and neither watches the other's room; Answer choices: (A) 128, (B) 144, (C) 152, (D) 156, (E) 168
Unknowns: The number of valid ordered (Alpha-room, Beta-room) placements
Understand
Restated: On a $4 \times 4$ grid of $16$ museum rooms, place two distinguishable drones Alpha and Beta — one drone per room, two different rooms — so that neither drone is in a room watched by the other. "Watched" means horizontally, vertically, or diagonally one step away. Count the number of ordered (Alpha-room, Beta-room) schedules that satisfy this.
Givens: Grid: $4 \times 4 = 16$ display rooms; Each drone watches every room one step away (up to $8$ rooms, like a chess-king move); Alpha and Beta carry different sensors, so the assignment is ordered (Alpha at room $A$, Beta at room $B$ differs from the swap); The two drones occupy different rooms, and neither watches the other's room; Answer choices: (A) 128, (B) 144, (C) 152, (D) 156, (E) 168
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #2 Make a Systematic List, #3 Eliminate Possibilities
Everything sits on a tiny $4 \times 4$ picture, so first use Tool #1 — **draw the grid** and color the rooms by where they sit. The picture instantly shows the $16$ rooms come in just three flavors by location: **corner** (only $3$ neighboring rooms), **side** (along an edge but not a corner — $5$ neighbors), and **interior** (any of the four innermost rooms — $8$ neighbors). Tool #7 (identify subproblems) splits the big count into one subproblem per flavor — Alpha sits in a corner / on a side / in the interior — and inside each subproblem Tool #2 (systematic listing via the multiplication rule) counts Beta's legal rooms. Finally Tool #3 (eliminate possibilities) checks that the total matches one of (A)-(E). Visual + counting tools are plenty for a $4 \times 4$ board — no algebra needed.
Execute — Answer: D
K.G.A.1 Step 1 - Sketch the $4 \times 4$ grid and sort the $16$ rooms by location.
- The **4 corner rooms** sit at the four corners of the grid; the **8 side rooms** are the non-corner rooms along the four outer edges (2 per edge × 4 edges); the **4 interior rooms** are the inner $2 \times 2$ block in the middle.
- The three flavors partition the grid: $4 + 8 + 4 = 16$, every room classified exactly once.
💡 Sorting rooms by position words like "corner", "side", and "inside" is the Kindergarten position-vocabulary skill.
K.G.A.1 Step 2 - Read off how many rooms a drone watches from each flavor.
- A drone in a **corner** has only $3$ neighboring rooms inside the grid (the two adjacent corners are off-board, so just the right-of, below, and diagonal).
- A drone on a **side** has $5$ neighbors (three along its own edge plus two reaching one row inward).
- A drone in the **interior** has the full $8$ neighbors.
- The three watching-counts $3, 5, 8$ are the key numbers per flavor.
💡 Counting the up-to-$8$ neighbors around a single room is still a Kindergarten-level adjacency activity.
3.OA.A.1 Step 3 - Break the big problem ("total schedules") into **3 subproblems** by where Alpha sits: ① corner, ② side, ③ interior.
- For each subproblem the rooms Beta may occupy are $16 - 1 - (\text{watches})$: subtract Alpha's own room (1) and the rooms Alpha watches.
- By the multiplication rule the contribution of that subproblem is $(\text{Alpha rooms}) \times (\text{Beta rooms})$, and the three contributions get added because the three flavors of Alpha-rooms are disjoint.
💡 Reading "$A$ choices, each with $B$ choices, gives $A \times B$ total" is the Grade 3 product-of-equal-groups meaning of multiplication.
3.OA.C.7 Step 4 - **Subproblem ①: Alpha in a corner.** Alpha rooms: $4$.
- The corner drone blocks its own room plus the $3$ it watches, so $1 + 3 = 4$ rooms are off-limits to Beta.
- Beta picks from $16 - 4 = 12$ rooms.
- Contribution: $4 \times 12 = 48$.
💡 $4 \times 12 = 48$ is fluent multiplication within $100$, a Grade 3 standard.
3.OA.C.7 Step 5 - **Subproblem ②: Alpha on a side room.** Alpha rooms: $8$.
- The side drone blocks $1 + 5 = 6$ rooms, leaving Beta with $16 - 6 = 10$.
- Contribution: $8 \times 10 = 80$.
💡 $8 \times 10 = 80$ is a basic Grade 3 multiplication-within-$100$ fact.
3.OA.C.7 Step 6 - **Subproblem ③: Alpha in the interior.** Alpha rooms: $4$.
- The interior drone blocks $1 + 8 = 9$ rooms, leaving Beta with $16 - 9 = 7$.
- Contribution: $4 \times 7 = 28$.
💡 $4 \times 7 = 28$ is another Grade 3 multiplication fact.
3.NBT.A.2 Step 7 - Add the three subproblem contributions to get the total schedule count: $48 + 80 + 28 = 156$.
- The three Alpha-flavors partition the $16$ rooms with no overlap, so the addition rule applies directly.
💡 Adding three two-digit numbers like $48 + 80 + 28$ is squarely within Grade 3 add-within-$1000$ fluency.
2.NBT.A.4 Step 8 - Match $156$ against the answer choices.
- (A) $128$ is exactly the corner + side contributions $48 + 80$ (forgetting the interior subproblem).
- (B) $144$ and (C) $152$ are near-miss values that come from undercounting Beta-rooms in one flavor.
- (E) $168$ overshoots by $12$, the typical slip of counting interior watches as $6$ instead of $8$.
- Our total $156$ is exactly **(D)**.
💡 Choosing the matching three-digit number from a short list is a Grade 2 three-digit comparison skill.
K.G.A.1 Sketch the $4 \times 4$ grid and sort the $16$ rooms by location. The **4 corner K.G.A.1 Read off how many rooms a drone watches from each flavor. A drone in a **corner* 3.OA.A.1 Break the big problem ("total schedules") into **3 subproblems** by where Alpha 3.OA.C.7 **Subproblem ①: Alpha in a corner.** Alpha rooms: $4$. The corner drone blocks i 3.OA.C.7 **Subproblem ②: Alpha on a side room.** Alpha rooms: $8$. The side drone blocks 3.OA.C.7 **Subproblem ③: Alpha in the interior.** Alpha rooms: $4$. The interior drone bl 3.NBT.A.2 Add the three subproblem contributions to get the total schedule count: $48 + 80 2.NBT.A.4 Match $156$ against the answer choices. (A) $128$ is exactly the corner + side c Review
Reasonableness: Cross-check by counting the complement. Total ordered placements with Alpha and Beta in different rooms: $16 \times 15 = 240$. Now count ordered pairs of rooms that ARE adjacent. Horizontally or vertically adjacent unordered pairs on the $4 \times 4$ grid: $4 \cdot 3 + 4 \cdot 3 = 24$. Diagonally adjacent unordered pairs: $3 \cdot 3 \cdot 2 = 18$. Total adjacent unordered pairs: $24 + 18 = 42$; doubling for the ordered version gives $84$. Non-attacking ordered placements: $240 - 84 = 156$, matching our casework total exactly. Sanity-wise, the interior flavor giving the smallest contribution ($28$) makes sense because interior rooms watch the most other rooms.
Alternative: An alternative is Tool #16 (Change Focus / Count the Complement): compute the $240$ total ordered placements in different rooms, count the $84$ ordered adjacent pairs, and subtract to get $240 - 84 = 156$ (this is the cross-check above). For a young student, though, splitting Alpha's rooms into corner / side / interior and adding is more concrete, so the main solution uses #7 + #2.
CCSS standards used (min grade 3)
K.G.A.1Describe positions of objects using above, below, beside, in front of (Classifying the $16$ rooms into corner / side / interior flavors and counting each room's up-to-$8$ neighbors.)2.NBT.A.4Compare two three-digit numbers using symbols (Matching the computed total $156$ to the five three-digit answer choices to identify (D).)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Reading "(Alpha rooms) × (Beta rooms)" as the count per flavor — a product of equal-sized groups (the multiplication rule).)3.OA.C.7Fluently multiply and divide within 100 (Computing each flavor's contribution: $4 \times 12 = 48$, $8 \times 10 = 80$, and $4 \times 7 = 28$.)3.NBT.A.2Fluently add and subtract within 1000 (Adding the three flavor contributions $48 + 80 + 28 = 156$.)
⭐ This AMC 8 problem only needs Grade 3 multiplication and the "split into cases, then add" idea you already know!
⭐ This AMC 8 problem only needs Grade 3 multiplication and the "split into cases, then add" idea you already know!
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