Division With Remainders Answer Key - Complete Solutions for Grade 4 Students

Complete answer key with step-by-step solutions for division with remainders problems. Includes detailed explanations and verification methods.

Division With Remainders - Answer Key Summary
Key Definitions
Essential Terms:

Division: The process of splitting a number into equal parts or groups

Dividend: The number being divided (the larger number)

Divisor: The number that divides the dividend (the smaller number)

Quotient: The result of division (how many equal parts/groups)

Remainder: What's left over after division when the divisor doesn't divide evenly into the dividend

Division With Remainder: Division where the divisor doesn't divide evenly into the dividend, leaving a remainder

Answer Key: A guide that provides correct answers and solution methods for problems

Core Rules and Laws
Fundamental Rules:

• Division Relationship: Dividend = Divisor × Quotient + Remainder

• Remainder Rule: Remainder is always less than the divisor

• Division by 1: Any number divided by 1 equals itself

• Zero Division: 0 divided by any non-zero number equals 0

• Division by Zero: Undefined (impossible)

• Inverse Operation: Division is the inverse of multiplication

• Remainder Constraint: Remainder must always be smaller than the divisor

Step-by-Step Method
Division With Remainder Process:
  1. Divide: How many times does the divisor go into the dividend?
  2. Multiply: Multiply the quotient by the divisor
  3. Subtract: Subtract the result from the dividend
  4. Check: The result is the remainder (must be less than divisor)
  5. Verify: Use: Divisor × Quotient + Remainder = Dividend
Examples and Tips
Simple Example: 13 ÷ 4 = 3 remainder 1
With Remainder: 25 ÷ 6 = 4 remainder 1
Key Tip: The remainder is always less than the divisor!
Check Method: Always verify: Quotient × Divisor + Remainder = Original Number
Division With Remainders - Answer Key
Complete Solutions with Step-by-Step Explanations
Problem 1: 17 ÷ 5
Answer: 3 remainder 2
Step 1: 5 goes into 17 three times (5 × 3 = 15)
Step 2: 17 - 15 = 2
Step 3: 2 is less than 5, so 2 is the remainder
Step 4: Verification: 5 × 3 + 2 = 15 + 2 = 17 ✓
Check: 3 × 5 + 2 = 17 ✓
Problem 2: 23 ÷ 4
Answer: 5 remainder 3
Step 1: 4 goes into 23 five times (4 × 5 = 20)
Step 2: 23 - 20 = 3
Step 3: 3 is less than 4, so 3 is the remainder
Step 4: Verification: 4 × 5 + 3 = 20 + 3 = 23 ✓
Check: 5 × 4 + 3 = 23 ✓
Problem 3: 31 ÷ 6
Answer: 5 remainder 1
Step 1: 6 goes into 31 five times (6 × 5 = 30)
Step 2: 31 - 30 = 1
Step 3: 1 is less than 6, so 1 is the remainder
Step 4: Verification: 6 × 5 + 1 = 30 + 1 = 31 ✓
Check: 5 × 6 + 1 = 31 ✓
Problem 4: 42 ÷ 8
Answer: 5 remainder 2
Step 1: 8 goes into 42 five times (8 × 5 = 40)
Step 2: 42 - 40 = 2
Step 3: 2 is less than 8, so 2 is the remainder
Step 4: Verification: 8 × 5 + 2 = 40 + 2 = 42 ✓
Check: 5 × 8 + 2 = 42 ✓
Problem 5: 19 ÷ 3
Answer: 6 remainder 1
Step 1: 3 goes into 19 six times (3 × 6 = 18)
Step 2: 19 - 18 = 1
Step 3: 1 is less than 3, so 1 is the remainder
Step 4: Verification: 3 × 6 + 1 = 18 + 1 = 19 ✓
Check: 6 × 3 + 1 = 19 ✓
Common Mistakes and Corrections:
Mistake: 17 ÷ 5 = 2 remainder 7
Correction: Remainder must be less than divisor. 7 > 5, so incorrect.
Correct: 17 ÷ 5 = 3 remainder 2
Mistake: 23 ÷ 4 = 6 remainder 1
Correction: 4 × 6 = 24, which is greater than 23.
Correct: 23 ÷ 4 = 5 remainder 3
Advanced Division Solutions
Two-Digit and Three-Digit Division Problems
Problem 6: 84 ÷ 7
Answer: 12 remainder 0
Step 1: 7 goes into 8 once (7 × 1 = 7)
Step 2: 8 - 7 = 1, bring down 4 to make 14
Step 3: 7 goes into 14 twice (7 × 2 = 14)
Step 4: 14 - 14 = 0, no remainder
Step 5: Verification: 7 × 12 + 0 = 84 ✓
Check: 12 × 7 = 84 ✓
Problem 7: 91 ÷ 8
Answer: 11 remainder 3
Step 1: 8 goes into 9 once (8 × 1 = 8)
Step 2: 9 - 8 = 1, bring down 1 to make 11
Step 3: 8 goes into 11 once (8 × 1 = 8)
Step 4: 11 - 8 = 3, remainder is 3
Step 5: Verification: 8 × 11 + 3 = 88 + 3 = 91 ✓
Check: 11 × 8 + 3 = 91 ✓
Problem 8: 137 ÷ 9
Answer: 15 remainder 2
Step 1: 9 goes into 13 once (9 × 1 = 9)
Step 2: 13 - 9 = 4, bring down 7 to make 47
Step 3: 9 goes into 47 five times (9 × 5 = 45)
Step 4: 47 - 45 = 2, remainder is 2
Step 5: Verification: 9 × 15 + 2 = 135 + 2 = 137 ✓
Check: 15 × 9 + 2 = 137 ✓
Problem 9: 205 ÷ 6
Answer: 34 remainder 1
Step 1: 6 goes into 20 three times (6 × 3 = 18)
Step 2: 20 - 18 = 2, bring down 5 to make 25
Step 3: 6 goes into 25 four times (6 × 4 = 24)
Step 4: 25 - 24 = 1, remainder is 1
Step 5: Verification: 6 × 34 + 1 = 204 + 1 = 205 ✓
Check: 34 × 6 + 1 = 205 ✓
Problem 10: 318 ÷ 7
Answer: 45 remainder 3
Step 1: 7 goes into 31 four times (7 × 4 = 28)
Step 2: 31 - 28 = 3, bring down 8 to make 38
Step 3: 7 goes into 38 five times (7 × 5 = 35)
Step 4: 38 - 35 = 3, remainder is 3
Step 5: Verification: 7 × 45 + 3 = 315 + 3 = 318 ✓
Check: 45 × 7 + 3 = 318 ✓
Learning Checkpoints:
✓ Remainder is always less than the divisor
✓ Always verify your answer using: Divisor × Quotient + Remainder = Dividend
✓ Use multiplication facts to help with division
✓ Check your work step by step
✓ Practice regularly to improve speed and accuracy
11 Word Problem Solution
Problem 11
Sarah has 47 stickers and wants to put them in albums with 6 stickers per page. How many full pages can she fill and how many stickers will be left over?
Solution:

This is a division problem: 47 ÷ 6

6 goes into 47 seven times (6 × 7 = 42)

47 - 42 = 5

Answer: 7 full pages, 5 stickers left over

Step-by-step solution:

Step 1: Set up the division: 47 ÷ 6

Step 2: Find how many times 6 goes into 47

Step 3: 6 × 7 = 42, 6 × 8 = 48 (too big)

Step 4: So 6 goes into 47 seven times

Step 5: Calculate remainder: 47 - 42 = 5

Step 6: Verification: 6 × 7 + 5 = 42 + 5 = 47 ✓

Final answer:

Sarah can fill 7 full pages and will have 5 stickers left over.

Check: \(7 \times 6 + 5 = 42 + 5 = 47\) ✓

12 Advanced Word Problem Solution
Problem 12
A teacher has 127 students and wants to organize them in groups of 8. How many complete groups can be formed and how many students will be in a smaller group?
Solution:

This is a division problem: 127 ÷ 8

8 goes into 127 fifteen times (8 × 15 = 120)

127 - 120 = 7

Answer: 15 complete groups, 7 students in smaller group

Step-by-step solution:

Step 1: Set up the division: 127 ÷ 8

Step 2: Compare hundreds and tens: 8 goes into 12 once

Step 3: 12 - 8 = 4, bring down 7 to make 47

Step 4: 8 goes into 47 five times (8 × 5 = 40)

Step 5: 47 - 40 = 7, remainder is 7

Step 6: Verification: 8 × 15 + 7 = 120 + 7 = 127 ✓

Final answer:

The teacher can form 15 complete groups and will have 7 students in a smaller group.

Check: \(15 \times 8 + 7 = 120 + 7 = 127\) ✓

Questions & Answers

Question: How do I know if my answer to a division problem is correct?

Answer: There are several ways to check if your division answer is correct:

Method 1: Remainder Check

  • Make sure your remainder is smaller than your divisor
  • Example: In 17 ÷ 5, if you get remainder 7, that's wrong because 7 > 5

Method 2: Verification Formula

  • Check: Divisor × Quotient + Remainder = Original Dividend
  • Example: For 17 ÷ 5 = 3 remainder 2, check: 5 × 3 + 2 = 15 + 2 = 17 ✓

Method 3: Reasonableness Check

  • Does your answer make sense in the context of the problem?
  • Are you dividing the right numbers?

Always verify your answers! This helps you catch mistakes and builds confidence.

Question: What if I get confused about where to put the remainder? How do I know where it goes?

Answer: The remainder is what's left over after you finish the division process:

Where the remainder comes from:

  • After you multiply and subtract in the last step, whatever is left over is the remainder
  • Example: In 17 ÷ 5, you get 3 remainder 2 because 5 × 3 = 15, and 17 - 15 = 2
  • The 2 is what's left over after dividing 17 into groups of 5

How to write it:

  • Write: "quotient remainder remainder" (e.g., 3 remainder 2)
  • Or write: "quotient R remainder" (e.g., 3 R 2)
  • Some people write it as a mixed number: 3 2/5

Remember: The remainder is always what's left after equal distribution!

Question: Why do I sometimes get the wrong remainder? What causes mistakes?

Answer: Common mistakes that lead to wrong remainders:

Mistake 1: Calculation Errors

  • Multiplying incorrectly: 5 × 3 = 15 (not 14)
  • Subtracting incorrectly: 17 - 15 = 2 (not 3)
  • Forgetting to carry numbers in multi-digit problems

Mistake 2: Not Following the Process

  • Skipping steps in long division
  • Not bringing down the next digit properly
  • Not checking if the remainder is smaller than the divisor

Mistake 3: Confusion About the Remainder

  • Thinking the remainder can be larger than the divisor
  • Forgetting to write the remainder in the final answer
  • Mixing up the quotient and remainder

Tip: Always double-check your work by multiplying the divisor by the quotient and adding the remainder!