Visual walkthrough — Multiplication — tables (1–20), long multiplication, area model

We link the ideas we lean on: Place Value & Number Systems, the Distributive Law, and Addition — carrying and place value. This is the visual companion to the parent topic.

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Step 1 — A product IS an area

WHAT. Before any symbols, agree on one fact: if you draw a rectangle that is $b$ units wide and $a$ units tall, and you cut it into $1\times 1$ squares, the number of little squares is $a\times b$.

WHY this picture and not repeated addition? Repeated addition ($3+3+3+3$) is a list. A rectangle is a shape — and shapes can be cut into pieces whose areas add back up. That "cut and add" is the one move that makes big multiplication possible. So we trade the list for the rectangle on purpose.

PICTURE. Look at the figure: a rectangle $3$ tall, $5$ wide, filled with unit squares. Count them row by row ($5+5+5$) or all at once — it's $15$ either way, and $15=3\times 5$.

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Step 2 — Big numbers are built from place-value chunks

WHAT. Write each factor as a sum of its place values: $47=\underset{\text{4 tens}}{\underbrace{40}}+\underset{\text{7 ones}}{\underbrace{7}},23=\underset{\text{2 tens}}{\underbrace{20}}+\underset{\text{3 ones}}{\underbrace{3}}.$

WHY. We can't easily count a $47\times 23$ rectangle in one glance — it's huge. But we can count a rectangle whose sides are round, friendly pieces like $40$, $7$, $20$, $3$. So we split each side at its place-value seam. This split is exactly what Place Value & Number Systems is for: a digit's position tells you which chunk it is.

PICTURE. The figure shows one long side of length $47$ marked into a $40$ stretch (lavender) and a $7$ stretch (coral). The top side of length $23$ is marked into $20$ (mint) and $3$ (butter).

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Step 3 — The cuts make a 2×2 grid of tiles

WHAT. Draw the split on both sides at once. The two vertical-seam and two horizontal-seam lines carve the big rectangle into four smaller rectangles (tiles).

WHY. Each tile has round sides, so each tile's area is a table fact we already know. Four easy multiplications instead of one impossible one. This carving is the Distributive Law made visible — more on that in Step 4.

PICTURE. The figure labels each tile with its two side-lengths:

$\times$	width $20$	width $3$
height $40$	$40\times 20$	$40\times 3$
height $7$	$7\times 20$	$7\times 3$

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Step 4 — Fill in each tile (the four easy multiplications)

WHAT. Compute each tile's area:

$$\underset{\text{tens}\times \text{tens}}{\underbrace{40\times 20=800}},\underset{\text{tens}\times \text{ones}}{\underbrace{40\times 3=120}},\underset{\text{ones}\times \text{tens}}{\underbrace{7\times 20=140}},\underset{\text{ones}\times \text{ones}}{\underbrace{7\times 3=21}}.$$

WHY these are trivial. $40\times 20=4\times 2\times 10\times 10=8\times 100=800$: multiply the digit parts, then count the zeros. Every tile is just a small table fact ($4\times 2$, $4\times 3$, $7\times 2$, $7\times 3$) with zeros appended — that zero-appending is Place Value & Number Systems doing the heavy lifting.

PICTURE. The figure shows the four tiles now shaded and each stamped with its area, drawn roughly to scale so you see that $800$ is the biggest tile and $21$ the smallest.

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Step 5 — Add the tiles to get the answer

WHAT. Add all four areas: $800+140+120+21=1081.$

WHY grouped like this. Because the whole rectangle is the four tiles with no gaps and no overlaps (Step 2's guarantee), the total area — which equals $47\times 23$ — must be their sum. Addition here is ordinary column addition with carrying, exactly Addition — carrying and place value.

PICTURE. The figure stacks the four numbers and adds them, coloured to match their tiles, landing on $1081$.

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Step 6 — Long multiplication is this grid, folded into rows

WHAT. Long multiplication doesn't write four tiles. It writes two rows: one for $47\times 3$ and one for $47\times 20$. Compare:

$$\underset{\text{= }47\times 3\text{ = }141}{\underbrace{7\times 3+40\times 3}}(\text{bottom row of the grid}),\underset{\text{= }47\times 20\text{ = }940}{\underbrace{7\times 20+40\times 20}}(\text{top row of the grid}).$$

WHY the placeholder zero appears. The $2$ in $23$ is really $20$. Its whole row is $47\times 20=940$, which must end in a $0$. That trailing zero is not decoration — it's the tiles $7\times 20=140$ and $40\times 20=800$, both multiples of $10$, forced by Place Value & Number Systems.

PICTURE. The figure shows the grid on the left and the column algorithm on the right, with arrows: the two right tiles fuse into the "$141$" line, the two left tiles fuse into the "$940$" line.

     4 7
   × 2 3
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   1 4 1   ← 47 × 3   (= tiles 21 + 120)
 9 4 0     ← 47 × 20  (= tiles 140 + 800; the 0 is why)
 ------
 1 0 8 1

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Step 7 — Edge cases: zeros and single digits

WHAT. What if a factor has a $0$, or is a single digit? The rectangle still explains it.

Case A — a zero digit, e.g. $40\times 23$. Split $40=40+0$. The "$0$ chunk" is a tile of zero width — an invisible sliver of area $0$. So $40\times 23=40\times 20+40\times 3=800+120=920$; the ones-column tiles vanish.

Case B — single-digit multiplier, e.g. $47\times 6$. Now $6=6$ (no tens chunk). The grid has only one column of tiles: $7\times 6=42$ and $40\times 6=240$, summing to $282$. No placeholder zero is needed because there is no tens row.

WHY show these. A rule you can't apply to zeros or single digits isn't really understood. Here they're not exceptions — they're the grid with a tile of area $0$, or with one column missing.

PICTURE. The figure shows both: a grid whose ones-column has collapsed to zero width (Case A), and a one-column grid (Case B).

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The one-picture summary

Everything above compresses into a single labelled grid: split both sides by place value, read off four tiles, add them. Long multiplication just fuses the tiles into rows; the placeholder zero marks the tens row; single-digit and zero cases are the grid with a column removed or a tile of area zero.

Recall Feynman retelling — say it in plain words

Imagine tiling a big floor that's $47$ tiles one way and $23$ the other. Counting them all at once is scary. So you chalk two lines: split the $47$ side into a $40$ part and a $7$ part, split the $23$ side into a $20$ part and a $3$ part. Now the floor is four neat rooms: a big $40\times 20$ room ($800$ tiles), two medium rooms ($120$ and $140$), and a tiny $7\times 3$ room ($21$). Add the rooms: $800+140+120+21=1081$. Long multiplication is the exact same rooms, but bundled two rooms per line — the top and bottom rows of the grid. The little zero you write on the second line is there because that whole line came from the $20$ side, so every tile in it is a multiple of ten. Zeros just make a room infinitely thin (no tiles), and a single-digit multiplier just means one column of rooms. Same rectangle, every time.

Recall Rebuild the four tiles for

$47\times 23$ $40\times 20,40\times 3,7\times 20,7\times 3$ ::: $800,120,140,21$ — sum $1081$.

Recall Why does row two start with a zero?

Because that row is $47\times 20$, whose tiles ($800$ and $140$) are both multiples of ten. ::: The zero shifts the whole row into the tens column.

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Connections

• Place Value & Number Systems — why chunks like $40$ and $20$ multiply to a number ending in zeros.
• Distributive Law — the algebra of "cut the rectangle, add the tiles".
• Addition — carrying and place value — the final add-up step and carries.
• Algebra — Expanding Brackets — the same 2×2 grid with letters: $(x+a)(x+b)$.
• Division — inverse of multiplication — reading the rectangle backwards.
• Squares & Square Roots — the grid when both sides are equal.