Multiplication — tables (1–20), long multiplication, area model: Question bank

Words we'll lean on, all built in the parent note:

factor ::: one of the numbers being multiplied.
product ::: the answer of a multiplication.
partial product ::: the area of ONE cell in the area model — one place-value chunk times another.
place value ::: the idea that a digit's worth depends on its column (ones, tens, hundreds…), see Place Value & Number Systems.
distributive law ::: $a \times (b + c) = a \times b + a \times c$ , the engine, see Distributive Law.

True or false — justify

True or false:

a \times b

is always bigger than both

a

and

b

False. Only when both factors are

> 1

5 \times 1 = 5

(not bigger),

5 \times 0 = 0

(smaller than 5). "Multiplying makes things bigger" is a myth that dies the moment a factor is

0

1

True or false:

0 \times 7 = 7

False, it is

0

. "Zero copies of seven" means you took nothing — no groups at all — so the total is

0

. Zero groups can't add up to anything but zero.

True or false:

1 \times 7 = 7

because "one group of seven is seven".

True. One copy of

7

is just

7

itself; multiplying by

1

leaves a number unchanged (it is the identity for multiplication).

True or false:

6 \times 6 = 12

False.

6 \times 6

is six copies of six

= 36

, not

6 + 6 = 12

. Confusing "times" with "plus" is the classic trap — repeated addition means

6 + 6 + 6 + 6 + 6 + 6

, not one addition.

True or false:

7 \times 8 = 8 \times 7

, so I only ever need to learn half a times-table.

True. The dot grid (

7

rows of

8

8

rows of

7

, same dots) proves order doesn't change the count. Learn the upper triangle and mirror it (commutativity).

True or false:

a \times (b + c) = a \times b + c

False. The factor

a

must hit every term inside the bracket:

a \times b + a \times c

. In the area model both sub-cells have height

a

; you can't leave one cell short.

True or false: In

47 \times 23

, the second partial-product row

940

is a mistake because

47 \times 2 = 94

False,

940

is correct. The "2" in

23

means 2 tens, so that row is really

47 \times 20 = 940

. The trailing zero records the "tens" — it is not decoration.

True or false: To multiply by

10

you just "add a zero", so to multiply

2.5

10

you write

2.50

False. "Add a zero" is a whole-number shortcut that really means "shift every digit one place-value column left". For

2.5

that shift gives

25

, not

2.50

. Trust the place-value shift, not the surface trick.

True or false: If I double one factor, the product doubles.

True.

2 a \times b = 2 \times (a \times b)

— twice as many copies means twice the total. This is why the doubling trick (

\times 4 = \times 2

twice) works.

True or false: Making both factors bigger by 1 makes the product bigger by 1.

False.

(a + 1) (b + 1) = ab + a + b + 1

, so it grows by

a + b + 1

, not

1

. The area model shows a whole extra L-shaped strip is added, not a single cell.

Spot the error

A student writes: "

47 \times 6

7 \times 6 = 42

, so I write

42

under the ones." — what's wrong?

A column holds only one digit. The "

4

" in

42

4

tens and must carry to the tens column; only the "

2

" stays in the ones. Writing

42

stacked in one column double-counts place value.

A student computes

47 \times 23

and lines up

94

(from

47 \times 2

) directly under

141

in the ones place, getting

235

. Find the bug.

The missing placeholder zero. The "

2

" is

20

, so its row is

47 \times 20 = 940

and must sit one column left. Aligning it as plain

94

makes it

10 \times

too small.

A student says "

8 \times 5 = 40

, and

8 \times 50 = 45

, just add a

5

." Where did it break?

Multiplying by

50

\times 5

then

\times 10

40 \times 10 = 400

, not "add a 5". You scale the whole product by ten, you don't tack a digit onto it.

A student "checks"

9 \times 6

with the finger trick, bends finger 6, sees 5 fingers left and 4 right, and writes

45

— but claims the answer is

9 \times 6 = 45

. Is the trick wrong?

The trick is right and

9 \times 6 = 54

, but the student mislabelled: fingers left are tens (5 tens

= 50

) and fingers right are ones (

4

). Reading it as "45" swaps tens and ones. Answer:

54

A student expands

6 \times (10 - 1)

60 - 1 = 59

to find

6 \times 9

. Fix it.

Distribution applies to subtraction too:

6 \times 10 - 6 \times 1 = 60 - 6 = 54

. The

6

multiplies both pieces; only subtracting

1

instead of

6 \times 1

is the slip.

A student says

13 \times 12 = 12 \times 12 + 12 = 156

, then "so

13 \times 13 = 156 + 12 = 168

." Spot it.

Going from

\times 12

\times 13

adds one more group of 13, not 12:

13 \times 13 = 156 + 13 = 169

. (Which is also

1 3^{2}

— see Squares & Square Roots.) Always add a group equal to the number you are stepping across.

A student draws the area model for

47 \times 23

but splits only the width (

40 + 7

) and leaves the height as a single

23

, giving two cells. Is that valid?

Yes, valid and correct:

23 \times 40 + 23 \times 7 = 920 + 161 = 1081

. Splitting both sides into four cells is optional convenience, not a requirement — one split still uses the distributive law fully.

Why questions

Why does the placeholder zero appear in the second row of long multiplication, in plain words?

Because the digit doing the multiplying stands in the tens column, so its partial product is ten times bigger and belongs one column to the left. The zero is how "one column left" is written down.

Why can we "learn only half" of the times tables?

Because

a \times b = b \times a

: the grid of dots is the same whether you count rows-first or columns-first. Every fact above the diagonal has a mirror twin below it (commutativity).

Why is multiplying by

10

the same as shifting digits left one column?

Ten copies of a number push each digit into the next-higher place value (ones become tens, tens become hundreds…), leaving the ones column empty — which we fill with

0

. See Place Value & Number Systems.

Why is the area model "the same" as long multiplication?

Both are the distributive law applied to place-value chunks: long multiplication writes the partial products in columns, the area model draws them as rectangle cells. Same numbers, two costumes.

Why does the

\times 9

finger trick work?

Because

9 n = 10 (n - 1) + (10 - n)

. The fingers left of the bent one count

n - 1

tens; the fingers right count

10 - n

ones — exactly those two pieces.

Why must I add ALL the partial products, not just pick the biggest?

Because the rectangle's total area is the sum of every cell — leave one out and you've literally left a strip of the floor un-tiled. Each cell is a real chunk of the product.

Why does

\times 5 =

half of

\times 10

Because

5

is half of

10

: five copies is half as many as ten copies, so it's half the total.

8 \times 5 = (8 \times 10) \div 2 = 40

Why is division called the inverse of multiplication?

Because it undoes it: if

a \times b = p

, then

p \div b = a

recovers the missing factor. Multiplication builds the rectangle's area from its sides; division finds a side from the area — see Division — inverse of multiplication.

Edge cases

What is

0 \times 0

, and why?

0

. Zero copies of zero is nothing added to nothing — no groups, and even if there were, each holds nothing. The empty case still obeys the definition.

What is

n \times 1

for any whole number

n

n

itself. One copy of

n

is just

n

;

1

is the multiplicative identity, it leaves everything unchanged.

What is

n \times 0

for any

n

, and how does the area model show it?

0

, always. A rectangle with one side of length

0

has zero width — no rectangle exists, so no area. Any number of zero-length groups still totals nothing.

If a factor has a zero inside it, like

40 \times 6

, does anything special happen?

No new rule —

40 = 4 \times 10

, so

40 \times 6 = 24 \times 10 = 240

. The internal zero is just place value; multiply the non-zero part and re-attach the ten's shift.

Does

a \times b = b \times a

still hold when one factor is

0

1

Yes.

0 \times 5 = 5 \times 0 = 0

and

1 \times 5 = 5 \times 1 = 5

. Commutativity has no exceptions among whole numbers — the dot-grid argument works even for empty or single rows.

What happens to a product if you make one factor

0

The whole product collapses to

0

, no matter how large the other factor. One empty side wipes out the entire rectangle — this is the "zero property" and it's why a single zero factor is decisive.

(x + a) (x + b)

just an area model with letters?

Yes — the same four-cell picture, sides split as

(x + a)

and

(x + b)

, cells

x^{2}, b x, a x, ab

. This is exactly Algebra — Expanding Brackets; multiplication of numbers was the training wheels.

Connections

Distributive Law — the single engine behind every "why" above.
Place Value & Number Systems — the source of every placeholder-zero trap.
Addition — carrying and place value — carrying errors are really addition errors in disguise.
Division — inverse of multiplication — the "undo" perspective on these facts.
Squares & Square Roots — the $13 \times 13$ edge case lives here.
Algebra — Expanding Brackets — where the letter version of the area model goes next.