Chapter 4 · Dividing by a one-digit number
Lesson 4.4 · Dividing a two-digit number by a one-digit number (3)
Hua's Classroom for Shanghai Maths — One Lesson One Exercise (English edition, pages 71–72).
Lesson Alignment
How this Shanghai Maths lesson connects to learning goals, Common Core, and Eureka Math.
Learning Objectives
Students will be able to:
- 01Find the greatest number that fits inequalities with ×.
- 02Connect those bounds to division thinking.
- 03Solve equal-group competition / grouping stories.
Common Core Standards
| Code | Standard (focus for this lesson) |
|---|---|
| 3.OA.A.2 | Interpret whole-number quotients. |
| 3.OA.A.3 | Multiply/divide within 100 to solve word problems. |
| 3.OA.B.5 | Apply properties of operations. |
| 3.OA.B.6 | Understand division as unknown-factor. |
| 3.OA.C.7 | Fluently multiply and divide within 100. |
| 3.OA.D.8 | Two-step word problems. |
Eureka Math Correspondence
| Shanghai Maths | Focus | Eureka Math |
|---|---|---|
| Estimation | Bounds for factors | G3 M1 |
Core Concepts
Strengthen estimation: greatest/least digits in products and related division.
Greatest / Least Digits
Examples: $(\square) \times 4 < 26$, $6 \times (\square) < 35$ — find greatest possible.
Inequalities
Link to “how many groups fit without going over.”
Applications
Pupils equally divided into groups for a competition.
Key Terms & Definitions (Reference)
Quick glossary for this lesson.
| Term | Definition |
|---|---|
| Dividend | The number being divided. |
| Divisor | The number you divide by. |
| Quotient | The result of division. |
| Remainder | What is left after equal groups are made. |
| Relationship | dividend = divisor × quotient + remainder |
| Greatest possible | Largest digit/value that still satisfies a condition. |
Key Answers
Quick reference for this lesson’s exercise answers.
1. Greatest number in the brackets
- $(6)\times7<45$,\; $(6)\times4<26$,\; $68>9\times(7)$
- $3\times(9)<28$,\; $6\times(5)<35$,\; $47>(5)\times8$
2. Directly write the quotient
- $30\div6=5$,\; $19\div2=9\ldots1$,\; $38\div9=4\ldots2$,\; $20\div3=6\ldots2$,\; $27\div5=5\ldots2$
- $32\div6=5\ldots2$,\; $21\div4=5\ldots1$,\; $28\div8=3\ldots4$,\; $40\div7=5\ldots5$,\; $52\div6=8\ldots4$
- $51\div6=8\ldots3$,\; $38\div4=9\ldots2$,\; $53\div7=7\ldots4$,\; $80\div9=8\ldots8$,\; $46\div8=5\ldots6$
3. Column method
- $27\div5=5\ldots2$,\; $40\div6=6\ldots4$,\; $58\div8=7\ldots2$,\; $66\div9=7\ldots3$
- $54\div3=18$,\; $65\div5=13$,\; $64\div4=16$,\; $72\div6=12$
- $66\div5=13\ldots1$,\; $77\div6=12\ldots5$,\; $99\div8=12\ldots3$,\; $88\div7=12\ldots4$
4. Application problems
- $12$ pupils in each group
- $17$ bags;\; $3$ balls left
- $15$ coats
- $63$ peaches in total;\; $21$ each
5. Enhancement
Possible totals: $7$,\; $14$,\; $21$,\; $28$,\; $35$ (when each winner gets $q$ and $q$ are left, with $q<6$).