Division: The operation of splitting a number into equal parts
Dividend: The number being divided (the total amount)
Divisor: The number by which the dividend is divided (group size)
Quotient: The result showing how many times the divisor fits completely
Remainder: The leftover amount that doesn't fit into complete groups
Division Algorithm: Dividend = (Divisor × Quotient) + Remainder
• Remainder Rule: The remainder must always be smaller than the divisor
• Division Algorithm: Dividend = (Divisor × Quotient) + Remainder
• Verification: Always check your answer using the division algorithm
• Complete Groups: The quotient represents the number of complete groups
• Leftovers: The remainder represents what's left over after forming complete groups
- Identify: Find the dividend and divisor
- Estimate: Make a rough guess about the quotient
- Count: Determine how many times the divisor fits into the dividend
- Multiply: Calculate divisor × quotient
- Subtract: Find the remainder (dividend - product)
- Verify: Check: Dividend = (Divisor × Quotient) + Remainder
• Division shows how many complete groups can be made
• Remainder shows what's left over after complete groups
• Remainder is always smaller than the divisor
• Use the division algorithm to check your work
• Practice with larger numbers builds fluency
Section A: Basic Division Problems
4 × 1 = 4, 4 × 2 = 8, 4 × 3 = 12, 4 × 4 = 16, 4 × 5 = 20, 4 × 6 = 24
Since 24 > 23, 4 goes into 23 five times
Product: 4 × 5 = 20
Remainder: 23 - 20 = 3
Check: (4 × 5) + 3 = 20 + 3 = 23 ✓
5 × 1 = 5, 5 × 2 = 10, 5 × 3 = 15, 5 × 4 = 20, 5 × 5 = 25, 5 × 6 = 30, 5 × 7 = 35
Since 35 > 31, 5 goes into 31 six times
Product: 5 × 6 = 30
Remainder: 31 - 30 = 1
Check: (5 × 6) + 1 = 30 + 1 = 31 ✓
6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, 6 × 5 = 30, 6 × 6 = 36, 6 × 7 = 42, 6 × 8 = 48
Since 6 goes into 42 exactly seven times, remainder is 0
Check: (6 × 7) + 0 = 42 + 0 = 42 ✓
Section B: Word Problems
Division problem: 37 ÷ 6
6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, 6 × 5 = 30, 6 × 6 = 36, 6 × 7 = 42
Since 42 > 37, 6 goes into 37 six times
Each student gets 6 pencils, with 1 pencil left over
Check: (6 × 6) + 1 = 36 + 1 = 37 ✓
Section C: Challenging Problems
7 × 1 = 7, 7 × 2 = 14, 7 × 3 = 21, 7 × 4 = 28, 7 × 5 = 35, 7 × 6 = 42, 7 × 7 = 49, 7 × 8 = 56, 7 × 9 = 63
Since 63 > 58, 7 goes into 58 eight times
Product: 7 × 8 = 56
Remainder: 58 - 56 = 2
Check: (7 × 8) + 2 = 56 + 2 = 58 ✓
9 × 1 = 9, 9 × 2 = 18, 9 × 3 = 27, 9 × 4 = 36, 9 × 5 = 45, 9 × 6 = 54, 9 × 7 = 63, 9 × 8 = 72, 9 × 9 = 81, 9 × 10 = 90
Since 90 > 83, 9 goes into 83 nine times
Product: 9 × 9 = 81
Remainder: 83 - 81 = 2
Check: (9 × 9) + 2 = 81 + 2 = 83 ✓
Division problem: 147 ÷ 25
25 × 1 = 25, 25 × 2 = 50, 25 × 3 = 75, 25 × 4 = 100, 25 × 5 = 125, 25 × 6 = 150
Since 150 > 147, 25 goes into 147 five times with remainder 22
But 5 buses would only hold 125 students, so we need 6 buses to accommodate all 147 students
Check: (25 × 5) + 22 = 125 + 22 = 147 ✓
• Analyze multi-step problems
• Identify relevant information
• Apply division in context
• Interpret remainders meaningfully
• Use division algorithm to check
• Verify remainder is smaller than divisor
• Double-check calculations
• Ensure answer makes sense
• Practice with larger numbers
• Develop mental math skills
• Improve speed and accuracy
• Build confidence with remainders
• Estimate first, then calculate
• Use multiplication facts
• Count multiples systematically
• Draw models when needed
• Working with larger dividends and divisors
• Keeping track of multi-digit multiplication
• Ensuring accurate subtraction
• Interpreting remainders in complex contexts
• Managing longer calculation processes
• Use scratch paper for organization
• Double-check each step
• Practice multiplication facts
• Verify all answers
• Build confidence gradually