Multiplication and Division Facts to 100
Warm-up
Doubling chain: 3x2=6, 6x2=12, 12x2=24, 24x2=48. Each step doubles. Now: what is 6x4? (6x2x2 = 12x2 = 24.) 6x8? (6x2x2x2 = 48.) This shows that all powers-of-2 multiplications can be computed by repeated doubling from a known fact.
Explore
Array game: player 1 rolls two dice to get factors (e.g., 6 and 7). They colour a 6x7 array on grid paper and record the product. Player 2 does the same. After 5 rounds, each player calculates their total area. Highest total wins. Strategy: which arrays give the most product? This embeds multiplication in spatial and quantitative reasoning.
Formalize
Connect to division: I know 7x8=56. What division facts do I know? (56/7=8, 56/8=7.) Write the fact family: 7x8=56, 8x7=56, 56/7=8, 56/8=7. The four facts from one product. How does this help when I encounter a division problem?
Multiplication and Division Facts to 100
Hundred chart patterns: multiples of 3 highlighted in yellow; multiples of 6 in blue. Notice: every blue square is also yellow. Why? (All multiples of 6 are also multiples of 3.) This is divisibility reasoning embedded in a visual pattern.
Practice
Students play the array game for 10 rounds, record all products, and list the four fact-family members for 6 of the products. Exit ticket: use the doubling strategy to find 4x8.
Exit ticket
Students play the array game for 10 rounds, record all products, and list the four fact-family members for 6 of the products. Exit ticket: use the doubling strategy to find 4x8.
Route 1 — through 5s: 7 × 8 = (5 × 8) + (2 × 8) = 40 + 16 = 56. Fives and twos are the facts everyone owns; 7 splits into them.
Route 2 — doubling: 7 × 8 = double 7 × 4 = double 28 = 56. Any ×8 is three doublings of the number (7 → 14 → 28 → 56).
Route 3 — from the square: 8 × 8 = 64 is often known (squares stick); one fewer group of 8 → 64 − 8 = 56.
Step 4: The array picture behind Route 1: a 7-by-8 rectangle sliced into a 5-by-8 slab and a 2-by-8 slab. Slicing arrays IS the distributive property — no name needed yet, just the picture.
Why derivation beats pure memorization: a memorized 56 that slips returns as 54 and nobody notices; a derived 56 self-checks. Automaticity is still the goal — derivation is the safety net under it.
Step 1: Reread the question as multiplication with a hole: 7 × ▢ = 63.
Step 2: Retrieve from the sevens: 7 × 9 = 63. So 63 ÷ 7 = 9. Done — division facts are multiplication facts read backwards, all 100 of them.
Step 3: Practice the triangle family for 63: 7 × 9, 9 × 7, 63 ÷ 7, 63 ÷ 9. One retrieval, four facts.
Step 4: The near-miss drill that exposes fragile knowledge: 64 ÷ 7 — trap! No sevens fact lands on 64 (63 does; 70 does). Answer: "9 with 1 left over" — a preview of remainders, discovered through fact-family fluency rather than taught as a procedure.
Watch for: students scanning the ENTIRE times table hunting for 63. Fluency means going straight to the sevens (or nines) shelf. If they scan, drill the individual family, not more mixed sheets.
Step 1: List the nines and stare: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90.
Step 2: Harvest pattern one: digit sums — 1+8, 2+7, 3+6 … every product's digits sum to 9. Instant answer-checker.
Step 3: Harvest pattern two: 9 × n has tens digit n−1 (9 × 7 → tens digit 6, and 6+3=9 forces the ones: 63).
Step 4: Explain WHY (the part that turns tricks into math): 9 × 7 = 10 × 7 − 7 = 70 − 7 = 63. Every nine is a ten with one group refunded. The n−1 tens and the digits-sum-to-9 both fall out of that refund.
Step 5: Generalize the move: 9s lean on 10s; 4s lean on 2s (double twice); 6s lean on 5s (one more group); 7s lean on 5s and 2s. The whole table rests on 2, 5, 10, and squares.
Exit: each student writes their personal "last three stubborn facts" and the derivation rope they'll use for each until retrieval kicks in.