Multiplication and Division to Three Digits
Warm-up
Number talk: 234 x 5. Students solve mentally. Strategies: 200x5=1000, 30x5=150, 4x5=20; total 1170. Or: 234x5 = 234x10/2 = 2340/2 = 1170. Multiple valid approaches; all roads lead to 1170.
Explore
Feast planning problem: 312 guests, 6 per table. How many tables? What if the chef makes 3 pieces of bannock per person, and each tray holds 20 pieces? Solve both, interpreting remainders appropriately. Students present their full reasoning, not just the number.
Formalize
Area model for 47 x 23: draw a rectangle, 47 wide and 23 tall. Split: 47 = 40+7, 23 = 20+3. Four sections: 40x20=800, 40x3=120, 7x20=140, 7x3=21. Sum: 800+120+140+21=1081. Each section is a partial product.
Multiplication and Division to Three Digits
Remainder context comparison: 127/5 in three different problems. (1) 127 apples shared among 5 families: 25 each, 2 left over. (2) 127 cm of ribbon cut into 5-cm pieces: 25 pieces, 2 cm unused. (3) 127 students need groups of 5: 26 groups (round up because every student needs a group). Same calculation, three different answers in context.
Practice
Students solve 4 multiplication and 4 division problems with area models shown, interpreting any remainders in context. Exit ticket: 234 chairs need to fit into rows of 8. How many full rows, and how many extra chairs?
Exit ticket
Students solve 4 multiplication and 4 division problems with area models shown, interpreting any remainders in context. Exit ticket: 234 chairs need to fit into rows of 8. How many full rows, and how many extra chairs?
Estimate first: 36 × 24 ≈ 35 × 25 ≈ 875 (or 40 × 20 = 800). Expect mid-800s.
Way 1 — area model: split 36 = 30 + 6 and 24 = 20 + 4. Four rooms: 30×20 = 600, 30×4 = 120, 6×20 = 120, 6×4 = 24. Total: 864.
Way 2 — partial products (the model written as a column): 600 + 120 + 120 + 24, stacked and added: 864.
Way 3 — the compact algorithm: 36 × 4 = 144; 36 × 20 = 720 (the shifted row — its final 0 is the ×10, SAY so); 144 + 720 = 864.
Way 4 — the dodge: 36 × 24 = 36 × 25 − 36 = 900 − 36 = 864. Quarters of a hundred make 25 a friendly neighbour.
Debrief: all four agree at 864 ✓ estimate ✓. The model explains, the algorithm hurries, the dodge delights. Own all three registers.
The context: 987 pencils packed into boxes of 4… no — shared among 4 classrooms. (Sharing story chosen deliberately; watch the end.)
Step 1: Share the hundreds: 9 hundreds ÷ 4 → 2 hundreds each (8 used), 1 hundred left. Write 2, remainder 1 hundred.
Step 2: Trade: 1 hundred = 10 tens, joining the 8 tens → 18 tens. Share: 18 ÷ 4 = 4 tens each (16 used), 2 tens left.
Step 3: Trade: 2 tens = 20 ones, joining the 7 → 27 ones. Share: 27 ÷ 4 = 6 each, 3 left over.
Step 4: Read the result: 246 each, remainder 3. Every algorithm line was a share-then-trade; the "bring down" IS the trade.
Step 5: Interpret the 3 by story: pencils to classrooms → 3 spares in the cupboard (answer 246 r3). If instead the 987 were dollars split fairly → keep dividing into decimals: 246.75. If they were students into 4-person relay teams → 246 full teams (drop). Same division, three endings — the story always writes the last line.
The scenario: the class bakes 26 trays of 18 muffins. They keep 30 muffins for volunteers and sell the rest in bags of 6. How many bags?
Step 1: Plan the pipeline: total baked → subtract kept → divide by bag size. Estimate the end: 26×18 ≈ 25×20 = 500; minus 30 → 470; ÷6 → high 70s.
Step 2: Total: 26 × 18. Dodge via 26 × 18 = 26 × 20 − 26 × 2 = 520 − 52 = 468.
Step 3: After keeping 30: 468 − 30 = 438.
Step 4: Bags: 438 ÷ 6 = 73 exactly (6 × 73 = 438 ✓). Seventy-three bags — matching the estimate's high-70s call.
Step 5: The follow-up that tests understanding, not stamina: "muffins sell out; each bag went for 25 decision?" 30 muffins = 5 bags = $25. Yes — generosity has a price tag, visible only to those who finish the math.
Multi-step problems are won at the pipeline sketch. Require the plan BEFORE any digits move.