Solved Exercises on Division as Repeated Subtraction Grade 3 in Grade 3

Master division as repeated subtraction for grade 3: advanced practice, visual representation, and complex applications through these 3 detailed exercises.

Solution: Exercises 1 to 3
1 Advanced Repeated Subtraction Grade 3
Exercise 1
Work through the grade 3 example: 45 ÷ 5 = ?. Show the complete repeated subtraction process with advanced steps and verification.
Definition:

Repeated Subtraction Grade 3: An advanced demonstration of how division works by repeatedly subtracting the divisor from the dividend with multiple steps.

Division as Repeated Subtraction: Division can be thought of as how many times we can subtract the divisor from the dividend efficiently.

Grade 3 Steps: Each subtraction represents one complete step in the division process with careful tracking.

Grade 3 Concept:
This is how division works with larger numbers: we take away the same number over and over until we can't take it away anymore. The number of times we take it away is our answer!
Grade 3 method:
  1. Start with dividend: Begin with the total amount (45)
  2. Subtract divisor: Subtract the divisor (5) from the dividend
  3. Record each step: Write down each subtraction in order
  4. Count steps: Keep track of how many times you subtract
  5. Continue: Repeat until remainder is less than divisor
  6. Result: The count is the quotient (9)
45
(Dividend)
→
- 5
(Subtract divisor)
→
40
(Step 1)
→
- 5
(Subtract again)
→
35
(Step 2)
Step 1
45 - 5 = 40
Step 2
40 - 5 = 35
Step 3
35 - 5 = 30
Step 4
30 - 5 = 25
Step 5
25 - 5 = 20
Step 6
20 - 5 = 15
Step 7
15 - 5 = 10
Step 8
10 - 5 = 5
Step 9
5 - 5 = 0
Step 1
45 - 5 = 40
Step 2
40 - 5 = 35
Step 3
35 - 5 = 30
Step 4
30 - 5 = 25
Step 5
25 - 5 = 20
Step 6
20 - 5 = 15
Step 7
15 - 5 = 10
Step 8
10 - 5 = 5
Step 9
5 - 5 = 0
45
40
35
30
25
20
15
10
5
0
Start: 45
→
Subtract 5
→
Count: 9 times
→
Answer: 9
Step 1: Start with the dividend

We begin with 45 (the number being divided)

Step 2: Subtract the divisor (5) from the dividend

45 - 5 = 40 (First subtraction)

Step 3: Continue subtracting the divisor

40 - 5 = 35 (Second subtraction)
35 - 5 = 30 (Third subtraction)
30 - 5 = 25 (Fourth subtraction)
25 - 5 = 20 (Fifth subtraction)
20 - 5 = 15 (Sixth subtraction)
15 - 5 = 10 (Seventh subtraction)
10 - 5 = 5 (Eighth subtraction)
5 - 5 = 0 (Ninth subtraction)

Step 4: Count the number of subtractions

We subtracted 5 from 45 a total of 9 times to reach 0
So 45 ÷ 5 = 9

45 ÷ 5 = 9
Final answer:

45 ÷ 5 = 9 because we can subtract 5 from 45 exactly 9 times to reach 0
Verification: 9 × 5 = 45 ✓

Applied rules:

• Subtraction principle: Count how many times we subtract the divisor

• Division relationship: Dividend ÷ Divisor = Quotient

• Verification: Quotient × Divisor = Dividend

• Stop condition: Stop when remainder is less than divisor

Memory trick: "How many times can I take away the divisor?"
Common mistake: Forgetting to count the number of subtractions
2 Grade 3 Example with Remainder
Exercise 2
Work through the grade 3 example: 52 ÷ 7 = ?. Show how to handle remainders in the repeated subtraction process for advanced students.
Definition:

Remainder Grade 3: An advanced demonstration of division when the dividend cannot be completely divided by the divisor.

Grade 3 Stop Condition: When the result of subtraction is less than the divisor.

52
(Dividend)
→
- 7
(Subtract divisor)
→
= 7 remainder 3
(Quotient and remainder)
Grade 3 Pattern:
When we subtract 7 from 52 seven times, we get 3 left over. This 3 is our remainder!
Step 1
52 - 7 = 45
Step 2
45 - 7 = 38
Step 3
38 - 7 = 31
Step 4
31 - 7 = 24
Step 5
24 - 7 = 17
Step 6
17 - 7 = 10
Step 7
10 - 7 = 3
Step 1
52 - 7 = 45
Step 2
45 - 7 = 38
Step 3
38 - 7 = 31
Step 4
31 - 7 = 24
Step 5
24 - 7 = 17
Step 6
17 - 7 = 10
Step 7
10 - 7 = 3
52
45
38
31
24
17
10
3
Start: 52
→
Subtract 7
→
Count: 7 times
→
Remainder: 3
Step 1: Start with the dividend

We begin with 52 (the number being divided)

Step 2: Subtract the divisor (7) repeatedly

52 - 7 = 45 (First subtraction)
45 - 7 = 38 (Second subtraction)
38 - 7 = 31 (Third subtraction)
31 - 7 = 24 (Fourth subtraction)
24 - 7 = 17 (Fifth subtraction)
17 - 7 = 10 (Sixth subtraction)
10 - 7 = 3 (Seventh subtraction)

Step 3: Check if we can subtract again

Now we have 3, which is less than 7
We cannot subtract 7 from 3
So we stop here

Step 4: Count subtractions and identify remainder

We subtracted 7 from 52 exactly 7 times
We have 3 left over (remainder)
So 52 ÷ 7 = 7 remainder 3

52 ÷ 7 = 7 remainder 3
Final answer:

52 ÷ 7 = 7 remainder 3 because we can subtract 7 from 52 exactly 7 times with 3 left over
Verification: (7 × 7) + 3 = 49 + 3 = 52 ✓

Applied rules:

• Stop condition: Stop when remainder is less than divisor

• Remainder rule: Remainder is always smaller than divisor

• Division algorithm: Dividend = (Divisor × Quotient) + Remainder

• Verification: (Quotient × Divisor) + Remainder = Dividend

Tip: Always check that remainder < divisor
Pattern: Leftover amount becomes remainder when subtraction stops
3 Grade 3 Word Problem Example
Exercise 3
Sarah has 63 stickers. She wants to put them in albums with 8 stickers on each page. How many full pages can she fill? Use repeated subtraction to solve this grade 3 example.
Definition:

Grade 3 Word Problem: A complex real-world scenario demonstrating repeated subtraction in context.

Grade 3 Key Words: "Put in albums," "with 8 each," "how many full" indicate division.

63 stickers
(Total items)
→
÷ 8 per page
(Group size)
→
= 7 pages
(Number of groups)
Grade 3 Story:
Sarah has 63 stickers and puts 8 on each page. How many full pages can she make? We take away 8 at a time!
Step 1
63 - 8 = 55
Step 2
55 - 8 = 47
Step 3
47 - 8 = 39
Step 4
39 - 8 = 31
Step 5
31 - 8 = 23
Step 6
23 - 8 = 15
Step 7
15 - 8 = 7
Step 1
63 - 8 = 55
Step 2
55 - 8 = 47
Step 3
47 - 8 = 39
Step 4
39 - 8 = 31
Step 5
31 - 8 = 23
Step 6
23 - 8 = 15
Step 7
15 - 8 = 7
63
55
47
39
31
23
15
7
Total: 63 stickers
→
Put 8 on each page
→
Count: 7 full pages
→
Remainder: 7
Step 1: Identify what we know

Total stickers: 63
Stickers per page: 8
Need to find: Number of full pages

Step 2: Set up the repeated subtraction

We need to subtract 8 from 63 repeatedly
This shows how many full pages can be made

Step 3: Perform the subtractions

63 - 8 = 55 (1st page filled)
55 - 8 = 47 (2nd page filled)
47 - 8 = 39 (3rd page filled)
39 - 8 = 31 (4th page filled)
31 - 8 = 23 (5th page filled)
23 - 8 = 15 (6th page filled)
15 - 8 = 7 (7th page filled)

Step 4: Count and conclude

7 full pages can be made with 8 stickers each
63 ÷ 8 = 7 remainder 7
Verification: (7 pages × 8 stickers) + 7 remainder = 56 + 7 = 63 stickers total ✓

Sarah can fill 7 full pages
Final answer:

Sarah can fill 7 full pages because 63 ÷ 8 = 7 remainder 7
She can subtract 8 from 63 exactly 7 times to make 7 full pages

Applied rules:

• Word problem translation: Convert story to division problem

• Repeated subtraction: Count how many groups we can make

• Verification: Multiply quotient × divisor to check

• Real-world connection: Division helps organize items into groups

Tip: Look for "in each" or "per group" to identify division
Key words: Each, per, groups, albums, pages, boxes indicate division
Detailed Summary: Division as Repeated Subtraction Grade 3
\(\text{Dividend} ÷ \text{Divisor} = \text{Number of Times Subtraction Occurs}\)
Repeated Subtraction Formula
Division
↔
Repeated Subtraction
↔
Counting Groups
Core definitions:

Repeated Subtraction Grade 3: An advanced demonstration of division where we repeatedly subtract the divisor from the dividend and record each step.

Division as Repeated Subtraction: Division can be understood as counting how many times we can subtract the divisor from the dividend.

Grade 3 Steps: Each subtraction represents one complete step in the division process.

Remainder: The amount left over when division is not exact.

Grade 3 Explanation:
Division is like organizing! We take the same amount away over and over until we can't take it away anymore. The number of times we take it away is our answer! With larger numbers, we need to be more careful about counting.
Grade 3 repeated subtraction methodology:
  1. Start with dividend: Begin with the total amount being divided
  2. Subtract divisor: Subtract the divisor from the dividend
  3. Record step: Write down each subtraction in order
  4. Count steps: Keep track of how many times you subtract
  5. Continue: Repeat until remainder is less than divisor
  6. Result: The count is the quotient, leftover is remainder
Memory trick: "How many times can I take away the divisor?"
Quick check: Quotient × Divisor + Remainder = Dividend
Pattern recognition: Count the number of subtractions performed
Verification method: Always multiply back to check
Common misconceptions: Thinking subtraction is different from division - they're connected!
Key insight: Repeated subtraction provides visual understanding of division
Critical repeated subtraction rules:

• Subtraction principle: Count how many times we subtract the divisor

• Stop condition: Stop when remainder is less than divisor

• Remainder rule: Remainder is always smaller than divisor

• Verification: (Quotient × Divisor) + Remainder = Dividend

• Division relationship: Dividend ÷ Divisor = Quotient

Examples by Type:

No remainder: 45 ÷ 5 = 9 (subtract 5 from 45 exactly 9 times to reach 0)

With remainder: 52 ÷ 7 = 7 remainder 3 (subtract 7 from 52 exactly 7 times, leaving 3)

Word problem: 63 apples ÷ 8 per basket = 7 baskets remainder 7

Division as Repeated Subtraction Grade 3 Reference
Division Repeated Subtraction Steps Quotient Remainder
45 ÷ 5 45-5=40, 40-5=35, 35-5=30, 30-5=25, 25-5=20, 20-5=15, 15-5=10, 10-5=5, 5-5=0 9 steps 9 0
52 ÷ 7 52-7=45, 45-7=38, 38-7=31, 31-7=24, 24-7=17, 17-7=10, 10-7=3 7 steps 7 3
63 ÷ 8 63-8=55, 55-8=47, 47-8=39, 39-8=31, 31-8=23, 23-8=15, 15-8=7 7 steps 7 7
71 ÷ 9 71-9=62, 62-9=53, 53-9=44, 44-9=35, 35-9=26, 26-9=17, 17-9=8 7 steps 7 8
Key mathematical relationships:

Division algorithm: Dividend = (Divisor × Quotient) + Remainder.

Verification: (Quotient × Divisor) + Remainder = Dividend.

Subtraction connection: Quotient = Number of times divisor subtracted.

Remainder constraint: Remainder < Divisor.

Grade 3 repeated subtraction problem-solving strategies:
  1. Identify components: Find dividend and divisor
  2. Set up example: Begin with dividend
  3. Perform subtractions: Subtract divisor repeatedly and record each step
  4. Count results: Track number of subtractions
  5. Verify answer: Multiply to check your division
36 ÷ 4 = ?
9 steps
43 ÷ 6 = ?
7 steps, remainder 1
58 ÷ 7 = ?
8 steps, remainder 2
65 ÷ 8 = ? r?
8 steps, remainder 1
Practice tip: Write out each subtraction step to visualize the process
Memory aid: Count the number of subtraction operations
Important note: Repeated subtraction builds conceptual understanding of division
Learning tip: Practice with larger numbers to build confidence

Division as Repeated Subtraction Grade 3 Guide

📊
Repeated Subtraction Grade 3 Steps
1
Start with Dividend
2
Subtract Divisor
3
Record Step
4
Count Steps
5
Stop When Done
Grade 3 Keywords

• How many times can I subtract?

• Take away repeatedly

  • Count the subtractions

    • • Record each step

    • Stop when remainder is too small

    Practice Makes Perfect! 🌟
    Advanced Grade 3 Repeated Subtraction Concepts
    \(\text{Dividend} = (\text{Divisor} × \text{Quotient}) + \text{Remainder}\)
    Division Algorithm
    Grade 3 Thinking:
    Division with larger numbers requires more careful tracking of steps. Each subtraction is still the same process, just with more steps!
    Advanced concepts:

    Division Algorithm: The fundamental relationship: Dividend = (Divisor × Quotient) + Remainder.

    Even Division: When remainder is zero, division is exact.

    Odd Division: When remainder exists, division is not exact.

    Subtraction Patterns: Recognizing when division will have remainders.

    Advanced grade 3 repeated subtraction approach:
    1. Pattern recognition: Identifying when remainders occur
    2. Verification techniques: Multiple ways to check answers
    3. Contextual understanding: Real-world implications of remainders
    4. Relationship analysis: How parts affect each other
    5. Problem extension: What to do with remainders
    Special Division Cases:

    Division by 1: Any number ÷ 1 = itself (subtract 1 n times)

    Division by itself: Any number ÷ itself = 1 (subtract n once)

    Zero divided: 0 ÷ any number = 0 (subtract 0 times)

    Division by zero: Undefined (cannot subtract 0)

    Advanced tip: In word problems, remainders may need special handling
    Pattern recognition: Odd dividends divided by 2 always have remainder 1
    Verification method: Use the division algorithm to check
    Contextual awareness: Sometimes remainders need rounding up
    Common errors: Forgetting to count subtractions, miscalculating steps.
    Advanced insight: Repeated subtraction connects to multiplication and fact families.
    Advanced repeated subtraction principles:

    • Division relationship: Dividend = (Divisor × Quotient) + Remainder

    • Remainder bounds: 0 ≤ Remainder < Divisor

    • Exact division: Remainder = 0 when dividend is multiple of divisor

    • Subtraction principle: Quotient = Number of times divisor subtracted

    • Zero division: Division by zero is undefined

    • One division: Any number ÷ 1 = itself

    Questions & Answers

    Question: I get confused about when to stop subtracting in repeated subtraction for grade 3. How do I know?

    Answer: Here's when to stop:

    • Exact division: Stop when you reach exactly 0
    • With remainder: Stop when the number is smaller than the divisor
    • Example: For 52 ÷ 7, stop at 3 because 3 < 7
    • Never continue: If you can't subtract the divisor anymore

    The number left over becomes your remainder!

    Question: My child finds repeated subtraction for grade 3 slow with larger numbers. Should they learn this method?

    Answer: Yes, it's very important! Here's why:

    • Builds understanding: Shows what division really means
    • Visual connection: Helps connect subtraction to division
    • Foundation: Prepares for more advanced division methods
    • Conceptual clarity: Ensures understanding, not just memorization

    Once they understand the concept, they can move to faster methods!

    Question: What if I subtract the divisor one more time when the remainder is smaller in the grade 3 example?

    Answer: You should NOT subtract anymore when the remainder is smaller than the divisor! Here's why:

    • Rule: The remainder must always be smaller than the divisor
    • Example: If you have 3 left and divisor is 7, you can't subtract 7 from 3
    • Result: You would get a negative number, which is wrong
    • Correct: Stop when remainder < divisor

    The remainder is what's left after you can't subtract anymore!