3A · Chapter 5 · Lesson 2
Multiplication Without Regrouping
Multiply each place-value part and combine without exchanges.
Today's challenge
Calculate 213 × 3 using partial products. Try your own method, then compare it with a model.
What you'll learn
- Multiply each place-value part and combine without exchanges.
- Explain the connection between a diagram and a number sentence.
- Solve a fresh application without step-by-step hints.
Before this lesson
Place-value decomposition.
Prerequisite competency: 3A-C05-L01
Entry · 2 questions
Warm-up: prerequisite check
Try on paper before continuing. If the questions are hard, review the prerequisite concepts.
- E1
Find 5 × 500; explain the units of the product.
Show answer and reasoning
Answer: 2500
500 is 5 groups of 100; 5×5=25 groups of 100 equals 2,500.
- E2
Find 6 × 80; explain the units of the product.
Show answer and reasoning
Answer: 480
80 is 8 groups of 10; 6×8=48 groups of 10 equals 480.
Learn the concept
Words to know
- partial product
- a product of one place-value part
- regrouping
- exchange between place-value units
Multiply each place-value part and combine without exchanges.
Use a place-value chart or partial products. Multiply each digit as the number of its place units and recombine. Each column stays below ten units here.
Why this works
To find 23 × 3, think of 23 as two tens and three ones. Three groups of two tens make six tens; three groups of three ones make nine ones. Together they make 69. Because neither column reaches ten units, no exchange is needed.
Check the meaning
Estimate to the nearest ten as a quick check on the product size.
Concrete → Pictorial → Abstract
- Build: Use counters, grid paper, or labeled bundles to represent the story or quantity.
- Draw: Sketch an array, place-value exchange, or equal-unit bar; label what is known.
- Calculate: Write an equation matching the representation and solve.
- Verify: Use an inverse operation or a second strategy to check.
For advanced students: if you can already explain this concept, start with the independent or challenge set.
Worked examples and immediate checks
Explain each step aloud or in writing. Work the fresh question before revealing its solution.
Worked example 1
Reason and verify
Calculate 213 × 3 using partial products.
- Use a place-value chart or partial products. Multiply each digit as the number of its place units and recombine. Each column stays below ten units here.
- Split 213 by place value; multiply each part by 3 and add. Total is 639.
- Estimate to the nearest ten as a quick check on the product size.
Result: 639
Understanding check 1 — your turn
Calculate 331 × 3 using partial products.
Reveal reasoning
Answer: 993 — Split 331 by place value; multiply each part by 3 and add. Total is 993.
Worked example 2
Reason and verify
Calculate 223 × 2 using partial products.
- Use a place-value chart or partial products. Multiply each digit as the number of its place units and recombine. Each column stays below ten units here.
- Split 223 by place value; multiply each part by 2 and add. Total is 446.
- Estimate to the nearest ten as a quick check on the product size.
Result: 446
Understanding check 2 — your turn
Calculate 311 × 2 using partial products.
Reveal reasoning
Answer: 622 — Split 311 by place value; multiply each part by 2 and add. Total is 622.
Worked example 3
Reason and verify
Calculate 221 × 3 using partial products.
- Use a place-value chart or partial products. Multiply each digit as the number of its place units and recombine. Each column stays below ten units here.
- Split 221 by place value; multiply each part by 3 and add. Total is 663.
- Estimate to the nearest ten as a quick check on the product size.
Result: 663
Understanding check 3 — your turn
Calculate 311 × 3 using partial products.
Reveal reasoning
Answer: 933 — Split 311 by place value; multiply each part by 3 and add. Total is 933.
Guided · 4 questions
Guided practice
Practice with the model visible. Explain the units at each step.
- G1
Calculate 31 × 2 using partial products.
Show answer and reasoning
Answer: 62
Split 31 by place value; multiply each part by 2 and add. Total is 62.
- G2
Calculate 31 × 3 using partial products.
Show answer and reasoning
Answer: 93
Split 31 by place value; multiply each part by 3 and add. Total is 93.
- G3
Calculate 11 × 2 using partial products.
Show answer and reasoning
Answer: 22
Split 11 by place value; multiply each part by 2 and add. Total is 22.
- G4
Calculate 11 × 3 using partial products.
Show answer and reasoning
Answer: 33
Split 11 by place value; multiply each part by 3 and add. Total is 33.
Independent · 8 questions
Independent practice
Work without relying on the model or checking answers prematurely.
- I1
Calculate 111 × 2 using partial products.
Show answer and reasoning
Answer: 222
Split 111 by place value; multiply each part by 2 and add. Total is 222.
- I2
Calculate 112 × 3 using partial products.
Show answer and reasoning
Answer: 336
Split 112 by place value; multiply each part by 3 and add. Total is 336.
- I3
Calculate 122 × 2 using partial products.
Show answer and reasoning
Answer: 244
Split 122 by place value; multiply each part by 2 and add. Total is 244.
- I4
Calculate 122 × 3 using partial products.
Show answer and reasoning
Answer: 366
Split 122 by place value; multiply each part by 3 and add. Total is 366.
- I5
Calculate 232 × 2 using partial products.
Show answer and reasoning
Answer: 464
Split 232 by place value; multiply each part by 2 and add. Total is 464.
- I6
Calculate 232 × 3 using partial products.
Show answer and reasoning
Answer: 696
Split 232 by place value; multiply each part by 3 and add. Total is 696.
- I7
Calculate 212 × 2 using partial products.
Show answer and reasoning
Answer: 424
Split 212 by place value; multiply each part by 2 and add. Total is 424.
- I8
Calculate 212 × 3 using partial products.
Show answer and reasoning
Answer: 636
Split 212 by place value; multiply each part by 3 and add. Total is 636.
Challenge · 2 questions
Challenge practice
Solve original application and explanation problems. Draw and defend a useful model.
- C1
Explain why the 2 in 23 × 3 contributes 60 rather than 6.
Show answer and reasoning
Answer: The 2 represents 20
3×20=60 and 3×3=9, so the product is 69.
- C2
Which factor could give a product of 96 when multiplied by 32? Explain by partial products.
Show answer and reasoning
Answer: 3
32×3 = (30×3)+(2×3)=90+6=96; both components do not require regrouping.
Exit · 3 questions
Exit check
Solve independently; revisit the skill if an important explanation is missing.
- E1
Calculate 33 × 2 using partial products.
Show answer and reasoning
Answer: 66
Split 33 by place value; multiply each part by 2 and add. Total is 66.
- E2
Calculate 33 × 3 using partial products.
Show answer and reasoning
Answer: 99
Split 33 by place value; multiply each part by 3 and add. Total is 99.
- E3
Calculate 13 × 2 using partial products.
Show answer and reasoning
Answer: 26
Split 13 by place value; multiply each part by 2 and add. Total is 26.