Division Problems That Have Remainders - Complete Guide for Grade 4 Students

Master division problems with remainders: understanding remainders, long division method, quotients, and step-by-step solutions through comprehensive exercises and detailed explanations.

Division Problems With Remainders - 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

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

Place Value: Position of digits (hundreds, tens, ones) that determines their value

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

• Exact vs Remainder: If remainder is 0, division is exact; otherwise, there's a remainder

Step-by-Step Method
Long Division Process With Remainders:
  1. Divide: How many times does the divisor go into the current number?
  2. Multiply: Multiply the quotient digit by the divisor
  3. Subtract: Subtract the result from the current number
  4. Bring down: Bring down the next digit
  5. Repeat: Continue until all digits are processed
  6. Check remainder: If the final result is greater than 0, that's the remainder
  7. 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
Solution: Exercises 1 to 3
1 Basic Division With Remainder
Exercise 1
Find the quotient and remainder of:
\(17 \div 5\)
Understanding the Problem:

This is a basic division problem where 5 doesn't divide evenly into 17, so there will be a remainder

Solution Approach:
  1. Find how many times 5 goes into 17
  2. 5 × 3 = 15, 5 × 4 = 20 (too big)
  3. So 5 goes into 17 three times with some left over
  4. 17 - 15 = 2, so remainder is 2
Expression
\(17 \div 5\)
Division
\(3\) R\(2\)
Step 1: Set up the division

We need to divide 17 by 5

17 ÷ 5 = ?

Step 2: Find how many times 5 goes into 17

Think: 5 × ? = how close to 17 without going over

5 × 3 = 15, 5 × 4 = 20 (too big)

So 5 goes into 17 three times

Step 3: Calculate the remainder

Multiply: 3 × 5 = 15

Subtract: 17 - 15 = 2

So the remainder is 2

Step 4: Check if remainder is valid

Is 2 less than 5? Yes, so 2 is our remainder

\(17 \div 5 = 3\) remainder \(2\)
Final answer:

\(17 \div 5 = 3\) remainder \(2\)

Check: \(3 \times 5 + 2 = 15 + 2 = 17\) ✓

Applied rules:

• Remainder constraint: Remainder (2) is less than divisor (5)

• Division relationship: Dividend = Divisor × Quotient + Remainder

• Multiplication check: 3 × 5 + 2 = 17 ✓

2 Two-Digit Division With Remainder
Exercise 2
Find the quotient and remainder of:
\(47 \div 6\)
Understanding the Problem:

This 2-digit division will have a remainder since 6 doesn't divide evenly into 47

Expression
\(47 \div 6\)
Division
\(7\) R\(5\)
Step 1: Set up the division

We need to divide 47 by 6

47 ÷ 6 = ?

Step 2: Find how many times 6 goes into 47

Think: 6 × ? = how close to 47 without going over

6 × 7 = 42, 6 × 8 = 48 (too big)

So 6 goes into 47 seven times

Step 3: Calculate the remainder

Multiply: 7 × 6 = 42

Subtract: 47 - 42 = 5

So the remainder is 5

Step 4: Check if remainder is valid

Is 5 less than 6? Yes, so 5 is our remainder

Step 5: Verify the answer

Check: 7 × 6 + 5 = 42 + 5 = 47 ✓

\(47 \div 6 = 7\) remainder \(5\)
Final answer:

\(47 \div 6 = 7\) remainder \(5\)

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

Applied rules:

• Division algorithm: Dividend = Divisor × Quotient + Remainder

• Remainder constraint: Remainder must be less than divisor

• Estimation: Use multiplication facts to determine how many times divisor fits

3 Three-Digit Division With Remainder
Exercise 3
Find the quotient and remainder of:
\(134 \div 8\)
Understanding the Problem:

This 3-digit division will have a remainder. We need to find both the quotient and remainder.

Expression
\(134 \div 8\)
Division
\(16\) R\(6\)
Step 1: Set up the division

We need to divide 134 by 8

134 ÷ 8 = ?

Step 2: Compare hundreds and tens

How many times does 8 go into 13 (hundreds and tens)?

8 × 1 = 8, 8 × 2 = 16 (too big)

So 8 goes into 13 once (1)

Write 1 above the tens place

Step 3: Multiply and subtract

1 × 8 = 8

13 - 8 = 5

Step 4: Bring down the ones place

Bring down the 4

Now we have 54

Step 5: Divide the new number

How many times does 8 go into 54?

8 × 6 = 48, 8 × 7 = 56 (too big)

So 8 goes into 54 six times (6)

Write 6 above the ones place

Step 6: Multiply and subtract

6 × 8 = 48

54 - 48 = 6

6 is less than 8, so 6 is our remainder

\(134 \div 8 = 16\) remainder \(6\)
Final answer:

\(134 \div 8 = 16\) remainder \(6\)

Check: \(16 \times 8 + 6 = 128 + 6 = 134\) ✓

Applied rules:

• Long division: Divide, multiply, subtract, bring down, repeat

• Remainder constraint: Remainder (6) is less than divisor (8)

• Multiplication check: Verify result by multiplying quotient by divisor

Solution: Exercises 4 to 5
4 Word Problem With Remainder
Exercise 4
Sarah has 53 stickers and wants to put them in albums with 7 stickers per page. How many full pages can she fill and how many stickers will be left over?
Understanding the Problem:

This word problem requires division with remainder to find both complete groups and leftovers

Problem
\(53 \div 7\)
Division
\(7\) R\(4\)
Step 1: Identify the operation

Sarah has 53 stickers and puts 7 per page

To find full pages: 53 ÷ 7 = ?

Step 2: Find how many times 7 goes into 53

Think: 7 × ? = how close to 53 without going over

7 × 7 = 49, 7 × 8 = 56 (too big)

So 7 goes into 53 seven times

Step 3: Calculate the remainder

Multiply: 7 × 7 = 49

Subtract: 53 - 49 = 4

So the remainder is 4

Step 4: Interpret the results

Quotient (7): Number of full pages

Remainder (4): Stickers left over

Step 5: Verify the answer

Check: 7 × 7 + 4 = 49 + 4 = 53 ✓

7 full pages, 4 stickers left over
Final answer:

Sarah can fill 7 full pages and will have 4 stickers left over.

Check: \(7 \times 7 + 4 = 49 + 4 = 53\) ✓

Applied rules:

• Equal grouping: Division to organize items into equal groups

• Real-world context: Interpret the quotient and remainder in the problem's context

• Verification: Multiply quotient by divisor plus remainder to check

5 Complex Division With Remainder
Exercise 5
Solve: \(285 \div 12\) and verify your answer
Understanding the Problem:

This 3-digit by 2-digit division will have a remainder. We need to find the quotient and remainder, then verify our answer.

Expression
\(285 \div 12\)
Division
\(23\) R\(9\)
Step 1: Set up the division

We need to divide 285 by 12

285 ÷ 12 = ?

Step 2: Compare hundreds and tens

How many times does 12 go into 28 (hundreds and tens)?

12 × 2 = 24, 12 × 3 = 36 (too big)

So 12 goes into 28 two times (2)

Write 2 above the tens place

Step 3: Multiply and subtract

2 × 12 = 24

28 - 24 = 4

Step 4: Bring down the ones place

Bring down the 5

Now we have 45

Step 5: Divide the new number

How many times does 12 go into 45?

12 × 3 = 36, 12 × 4 = 48 (too big)

So 12 goes into 45 three times (3)

Write 3 above the ones place

Step 6: Multiply and subtract

3 × 12 = 36

45 - 36 = 9

9 is less than 12, so 9 is our remainder

Step 7: Verify the answer

Check: 23 × 12 + 9 = 276 + 9 = 285 ✓

Alternative check: 285 ÷ 12 = 23 remainder 9 ✓

\(285 \div 12 = 23\) remainder \(9\)
Final answer:

\(285 \div 12 = 23\) remainder \(9\)

Check: \(23 \times 12 + 9 = 276 + 9 = 285\) ✓

Applied rules:

• Division verification: Always check your answer

• Multiplication facts: Knowing 12 × 23 = 276 helps solve this

• Remainder constraint: Remainder (9) is less than divisor (12)

Questions & Answers

Question: I don't understand why we have remainders in division. What do they mean in real life?

Answer: Great question! Remainders represent "leftover" amounts when things can't be divided evenly. Here are real-life examples:

Sharing Example:

  • If you have 17 cookies and want to share equally among 5 friends
  • Each friend gets 3 cookies (15 total), and you have 2 cookies left over
  • The remainder (2) represents the cookies that can't be shared equally

Grouping Example:

  • If you have 47 students and buses that hold 6 students each
  • You need 7 full buses (42 students) and 5 students will ride in a partially full bus
  • The remainder (5) represents the extra students who don't fill a complete bus

Remainders are important because they tell us what's left after equal distribution!

Question: How do I know if my remainder is correct? Sometimes I get confused about whether it's right.

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

Rule 1: Remainder Must Be Smaller Than Divisor

  • If you get a remainder that's equal to or larger than your divisor, you made a mistake
  • Example: In 17 ÷ 5, if you got remainder 7, that's wrong because 7 > 5
  • The correct answer is 3 remainder 2 (because 2 < 5)

Rule 2: Verify Using the Division Formula

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

Rule 3: Reasonableness Check

  • Make sure your answer makes sense in the context of the problem
  • The remainder should represent what's left after equal distribution

Question: When do I know if a division problem will have a remainder or not?

Answer: You can predict if a division will have a remainder in a few ways:

Way 1: Look for Exact Multiples

  • If the dividend is an exact multiple of the divisor, there will be no remainder
  • Example: 24 ÷ 6 = 4 (no remainder) because 24 = 6 × 4
  • Example: 25 ÷ 6 = 4 remainder 1 because 25 is not a multiple of 6

Way 2: Trial Multiplication

  • Multiply the divisor by different numbers to see if you get the dividend exactly
  • For 17 ÷ 5: 5 × 3 = 15, 5 × 4 = 20
  • Since 17 is between 15 and 20, there will be a remainder

Way 3: During Calculation

  • When doing long division, if your final subtraction doesn't equal 0, that's your remainder
  • If it does equal 0, there's no remainder

Remember: Most division problems do have remainders unless the dividend is perfectly divisible by the divisor!