Multiplication — tables (1–20), long multiplication, area model: Worked examples

The scenario matrix

Before solving, let's map the territory. Every multiplication question you will meet in this chapter falls into one of these cells. Think of it as a checklist — by the end, every row has a worked example.

Cell	What makes it different	Where the trap hides
A. One-digit × one-digit	pure table recall, no place value	forgetting a fact → derive it
B. Two-digit × one-digit	one carry across columns	dropping the carry
C. Two-digit × two-digit	two partial products, one shift	missing the placeholder zero
D. Multiply by a power of ten	zeros appended, digits unchanged	adding wrong number of zeros
*E. A zero inside* the number**	e.g. $305 \times 7$ — a middle-column zero	forgetting the empty column still exists
F. Trailing-zero factor	e.g. $60 \times 40$ — strip zeros, reattach	losing count of stripped zeros
G. Trick-derived fact	×9, ×11, ×5, doubling from parent	over-cleverness → mis-derive
H. Word problem	you must build the multiplication	choosing wrong factors
I. Exam twist	reverse / find a missing digit	assuming it's a plain multiply

Zero itself is a degenerate input: any factor of $0$ makes the product $0$ , because "zero groups of anything" is nothing, and "anything, zero times" is also nothing. We flag it wherever it appears (cells D, E, F).

By the parent's tricks we mean the shortcut rules built in the parent note's "Derivation tricks" box: ×10 (append a zero), ×9 ( $n \times 10 - n$ ), ×11 (write the digit twice, for $n \leq 9$ ), ×5 (half of ×10), and doubling (×2 repeated). Cell G below drills three of them.

The examples

Cell A — one-digit × one-digit (pure recall, with a rescue)

Cell B — two-digit × one-digit (one carry)

Worked example Compute

$47 \times 6$ , watching the carry. Forecast: $47$ is nearly $50$ , and $50 \times 6 = 300$ , so expect a bit under $300$ , around $280$ .

Look at Figure s01 below — it draws $47$ as a rectangle of height $6$ , sliced into an orange tens block ( $40$ wide) and a teal ones block ( $7$ wide). The two block areas are the two partial products we are about to compute; their total area is the answer.

Figure s01 — the strip $47 \times 6$ . Orange block area $= 40 \times 6 = 240$ ; teal block area $= 7 \times 6 = 42$ ; together they tile the whole strip, $282$ .

Split $47 = 40 + 7$ . Why this step? Place value: $47$ is $40 + 7$ , and the distributive law lets us hit each piece with the $6$ — exactly the two blocks in Figure s01.
Ones (teal block): $7 \times 6 = 42$ . Write 2, carry 4. Why this step? $42 = 4$ tens $+ 2$ ones. Only one digit fits the ones column; the $4$ tens must travel to the tens column — that is what "carry" means.
Tens (orange block): $4 \times 6 = 24$ tens, plus the carried $4$ tens $= 28$ tens $= 280$ . Why this step? We finish the second block and fold in the carry so the $4$ tens from step 2 are not lost.
Total $= 280 + 2 = 282$ . Why this step? Adding the two block areas ( $240 + 42$ ) reassembles the whole strip — the distributive law's promise that the pieces sum to the product.

Verify: $40 \times 6 + 7 \times 6 = 240 + 42 = 282$ . ✓ Sits just under $300$ as forecast.

Cell C — two-digit × two-digit (two partial products, the shift)

Worked example Compute

$47 \times 23$ — the full area picture. Forecast: round to $50 \times 20 = 1000$ ; the true answer is a little more. Guess ~1050.

Figure s02 below splits both sides of the rectangle by place value, creating a $2 \times 2$ grid. Each coloured cell is one partial product; the whole rectangle's area is their sum.

Figure s02 — the $47 \times 23$ rectangle cut into four cells. Orange $800$ , teal $140$ , plum $120$ , ink $21$ ; the four areas tile the full rectangle, $1081$ .

Split both factors: $47 = 40 + 7$ , $23 = 20 + 3$ . Why this step? Two place-value splits → a $2 \times 2$ grid of easy multiplications, the four cells drawn in Figure s02.
Fill the four cells:
- $20 \times 40 = 800$
- $20 \times 7 = 140$
- $3 \times 40 = 120$
- $3 \times 7 = 21$ Why this step? Each cell is one chunk of one factor times one chunk of the other — all trivial once the zeros are visible.
Add: $800 + 140 + 120 + 21 = 1081$ . Why this step? Total area = sum of the four cell areas; the distributive law applied to both factors guarantees these four pieces rebuild the whole product.

Why the "shift" in column form? The top row $800 + 140 = 940 = 47 \times 20$ — it ends in a $0$ because everything there was multiplied by twenty, not two. That trailing zero is the placeholder.

Verify: column method — $47 \times 3 = 141$ , $47 \times 20 = 940$ , sum $141 + 940 = 1081$ . ✓ Just above the ~1050 forecast.

Cell D — multiplying by a power of ten (and by zero)

Cell E — a zero inside the number (empty middle column)

Worked example Compute

$305 \times 7$ . Forecast: near $300 \times 7 = 2100$ plus a touch more → ~2135.

The danger: the middle digit is $0$ , and beginners "skip" it and squash the columns. Figure s03 below lays out the three place-value columns and shows the carry from the ones landing inside the zero column.

Figure s03 — the columns of $305 \times 7$ . The plum arrow warns: the tens digit is $0$ , but the carried $3$ still lands here, so the column cannot be skipped. Answer $2135$ .

Split by place value: $305 = 300 + 0 + 5$ . Why this step? The hundreds, tens, and ones columns each still exist — the tens column just holds $0$ . Writing it out keeps the columns honest (see Figure s03).
Ones: $5 \times 7 = 35$ . Write 5, carry 3. Why this step? $35 = 3$ tens $+ 5$ ones; the $5$ stays, the $3$ tens carry left, exactly as in Cell B.
Tens: $0 \times 7 = 0$ , plus carried $3 = 3$ . Write 3. Why this step? Even though the tens digit is $0$ , the carry from the ones still lands here. Skipping the column would misplace everything to its left.
Hundreds: $3 \times 7 = 21$ . Write 21. Why this step? The last (leftmost) block has no further column to carry into, so its full value $21$ is written out.
Total $= 2135$ . Why this step? Reading the columns left to right gives $21 ∣ 3 ∣ 5 = 2135$ ; each column already holds a single digit, so no more carrying is needed.

Verify: $300 \times 7 + 5 \times 7 = 2100 + 35 = 2135$ . ✓ Matches forecast.

Cell F — both factors have trailing zeros

Cell G — trick-derived facts (×9, ×11 and ×5)

Worked example Compute

$8 \times 9$ , $7 \times 11$ and $8 \times 5$ using the parent's tricks (the ×9, ×11 and ×5 shortcut rules described in the scenario-matrix note above). Forecast: $8 \times 9$ is one group below $8 \times 10 = 80$ → ~72. $7 \times 11$ is one group above $7 \times 10 = 70$ → ~77. $8 \times 5$ is half of $8 \times 10 = 80$ → ~40.

×9 rule: $n \times 9 = n \times 10 - n$ . So $8 \times 9 = 80 - 8 = 72$ . Why this step? $9 = 10 - 1$ , so nine groups of $8$ = ten groups minus one group. Distributive law with a subtraction: $8 \times (10 - 1) = 80 - 8$ .
×11 rule (single digit): $n \times 11 = n \times 10 + n$ . So $7 \times 11 = 70 + 7 = 77$ — write the digit twice. Why this step? $11 = 10 + 1$ : ten groups plus one more group.
×5 rule: $n \times 5 = (n \times 10) \div 2$ . So $8 \times 5 = 80 \div 2 = 40$ . Why this step? $5$ is half of $10$ , so five groups is exactly half of ten groups — take the easy ×10 and halve it.

Verify: $8 \times 9$ : count nine eights or eight nines $\to 72$ ✓. $7 \times 11 = 77$ (double digit) ✓. $8 \times 5 = 40$ ✓.

Cell H — a word problem (you build the multiplication)

Worked example A classroom has

$23$ rows of chairs with $18$ chairs in each row. How many chairs? Forecast: $\approx 20 \times 20 = 400$ , so expect around $400$ – $420$ .

Translate to a product. "23 rows of 18" is $23$ copies of $18$ → $23 \times 18$ . Why this step? Equal groups (each row = 18 chairs) is exactly the definition of multiplication — repeated addition made fast.
Area model / split: $18 = 10 + 8$ , $23 = 20 + 3$ .
- $20 \times 10 = 200$ , $20 \times 8 = 160$ , $3 \times 10 = 30$ , $3 \times 8 = 24$ . Why this step? Splitting both factors by place value turns one hard product into four trivial cells (same four-cell idea as Figure s02).
Add: $200 + 160 + 30 + 24 = 414$ chairs. Why this step? The four cell areas tile the whole rectangle, so their sum is the total chair count — the distributive law applied to both factors.

Verify: column method — $23 \times 8 = 184$ , $23 \times 10 = 230$ , sum $184 + 230 = 414$ . ✓ Units are chairs (rows × chairs-per-row), and $414$ sits in the forecast band. ✓

Cell I — an exam twist (find the missing digit)

Which cell? — a quick decision map

Active recall

Recall Why does a middle zero (like in

$305$ ) still need attention? Because a carry from the column on its right can land there. $0 \times 7 = 0$ , but $0 + carried 3 = 3$ . Skipping the column mis-aligns every digit to its left.

Recall In

$60 \times 40$ , how many zeros get reattached and why? Two — one borrowed from each factor. Stripping gave $6 \times 4 = 24$ ; we removed $10 \times 10 = 100$ , so we multiply back by $100 \Rightarrow 2400$ .

Recall Turn "23 rows of 18 chairs" into a calculation and its answer.

$23 \times 18 = 414$ chairs (equal groups → multiplication).

Missing-digit rule

If product and one factor are known, the missing factor is product ÷ known factor.

8 \times 9

via the ×9 trick

80 - 8 = 72

8 \times 5

via the ×5 trick

half of

80 = 40

60 \times 40

answer

2400

305 \times 7

answer

2135

Connections

Parent topic
Distributive Law (the engine behind every cell above)
Place Value & Number Systems (why the placeholder zero and column shifts work)
Addition — carrying and place value (carries reduce to addition)
Division — inverse of multiplication (Cell I reverses a product)
Squares & Square Roots (near-square shortcuts speed recall)
Algebra — Expanding Brackets (the $2 \times 2$ grid becomes $(x + a) (x + b)$ )