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

What: $8\times 7$ and $7\times 8$ are the same product read two ways. Why: a grid of $8$ rows of $7$ dots has the same dots as $7$ rows of $8$ dots — you just turned your head. So $a\times b=b\times a$. Value: $8\times 7=56$. You only ever store this once.

$0\times 7$: "zero groups of seven" — you took no groups at all, so there's nothing to add up: $0\times 7=0$. $7\times 0$: "seven groups of zero" — you took seven groups, but each holds nothing: $0+0+0+0+0+0+0=0$. By commutativity this matches $0\times 7$. So any number times $0$ is $0$. $1\times 7$: "one group of seven" — a single group, nothing to add to it: $1\times 7=7$. $7\times 1$: "seven groups of one" — $1+1+1+1+1+1+1=7$. So any number times $1$ is itself. Why these matter: $0$ is the absorbing factor (it swallows everything to $0$) and $1$ is the identity factor (it leaves the number unchanged). Every long multiplication silently uses "×0 gives 0" and "×1 gives itself" inside its columns.

What: replace $9$ by $10-1$. Why: ×10 is trivial (append a zero) and subtracting is easy. $6\times 9=6\times 10-6=60-6=54$.

$14\times 10=140$, then halve: $140÷2=70$. So $14\times 5=70$. Why halving works: $5$ is exactly half of $10$, so five copies is half of ten copies.

Split $63=60+7$.

1. Ones: $3\times 7=21$. Write 1, carry 2 (that's 2 tens). Why: $21$ is $2$ tens $+1$ one; only the one digit fits this column.
2. Tens: $6\times 7=42$ tens, plus carried $2$ tens $=44$ tens. Why we can write both digits "44" at once now — and why no further carry: this is the last column (nothing sits to its left except empty space), so the tens-of-tens digit has an empty column waiting to receive it. The "$44$ tens" means $4$ tens ($40$) and $4$ hundreds ($400$); since there is no column beyond, both digits simply spill into the two open columns to the left with no overflow to carry onward. Contrast with step 1, where a full column did have a neighbour to carry into.
3. Result: 441. Check: $60\times 7+3\times 7=420+21=441$. ✓

Split $34=30+4$. Keep the digits in their place-value columns (Th · H · T · O, defined just above).

• Row 1 ($58\times 4$): $8\times 4=32$, write $2$ carry $3$; $5\times 4=20$, plus carried $3=23$. Row $1$ reads 232 (so $2$ in H, $3$ in T, $2$ in O).
• Row 2 ($58\times 30$): drop a placeholder $0$ in the ones (O) column first, then $58\times 3=174$. Row $2$ reads 1740 ($1$ in Th, $7$ in H, $4$ in T, $0$ in O).
• Add column by column: $232+1740=1972$.

Placeholder zero: the $3$ in $34$ is 3 tens, so its row is $58\times 30$, which must end in $0$. That $0$ sits in the O column, pushing $174$ one column left into T/H/Th — the reason this partial product is ten times bigger. Final check: $58\times 30+58\times 4=1740+232=1972$.

Split $47=40+7$ and $26=20+6$. The figure below draws the actual rectangle — this is the one place where a picture adds something the table cannot: it shows the four partial products as physical areas that tile one big rectangle with no gaps or overlaps, which is what makes the distributive law visible rather than just arithmetic.

Sum $=800+140+240+42=1222$. Match to long multiplication: the figure's top band ($800+140=940=47\times 20$) is exactly the first partial product; the bottom band ($240+42=282=47\times 6$) is the second. Reading the figure by bands recovers the two stacked rows; reading it by four cells recovers the table. Same maths, one drawn, one stacked.

Split $15=10+5$ (leave $16$ whole): $16\times 15=16\times 10+16\times 5=160+80=240.$ Why smart: ×10 (append zero) and ×5 (half of ×10) are the two easiest tricks, and $10+5=15$ exactly. Splitting the other factor ($16=10+6$) would work too but gives less-friendly pieces ($150+90$). Choose the split that lands on trivial operations.

The slip: they distributed $23$ over only the first term of $20+3$ and forgot the second factor still multiplies the $3$. Correct expansion: $23\times (20+3)=23\times 20+23\times 3=460+69=529$. Why $23\times 3$, not just $3$: every term inside the bracket is hit by the whole $23$ — draw the rectangle and both sub-cells have height $23$.

$99\times 47=(100-1)\times 47=4700-47=4653.$ Why faster: ×100 just appends two zeros, and one subtraction beats computing and aligning four partial products. Recognising a factor sits near a round number ($99\approx 100$) turns a big multiply into a shift-and-subtract.

Split $18=10+8$, $24=20+4$.

Total $=200+40+160+32=432$ chairs. Biggest block: the $10\times 20=200$ cell — that's the large corner rectangle of "tens of rows × tens of chairs", where most seats live. The tiny $8\times 4=32$ cell is the little corner.

(a) Near-square: $13\times 12=12\times 12+1\times 12=144+12=156$. Why $+12$ and not $+13$: going from $12$ groups of $12$ to $13$ groups of $12$ adds one more group of $12$. Watch the direction! (b) Area model: $13=10+3$, $12=10+2$: cells $100,20,30,6$. Sum $=100+20+30+6=156$. Agree: both give $156$. ✓

Bound it, don't guess. Replace each factor by a round number below and above it, keeping both factors on the same side:

• Lower bound: round both factors down — $30\times 10=300$. The true product can't be smaller (both factors are at least this big).
• Upper bound: round both factors up — $40\times 20=800$. The true product can't be larger (both factors are at most this big). So $300\le 34\times 12\le 800$ — guaranteed, before any real work. Tighter forecast: now keep $34$ exact and squeeze $12$ between two easy multipliers. A floor comes from the nearest round number below $12$, namely $10$: $34\times 10=340$. A ceiling comes from an easy number above $12$ — I pick $15$ because $34\times 15$ is quick (it's $34\times 10+34\times 5=340+170=510$) and $15$ still safely sits above $12$; a nearer choice like $13$ or $14$ would be tighter but $\times 13$/$\times 14$ aren't "instant" the way $\times 15$ (ten-plus-half-of-ten) is. So the trade is tightness vs. ease, and $15$ buys a still-useful ceiling with a one-line computation. Hence most likely $340$–$510$. Exact: $34\times 12=34\times 10+34\times 2=340+68=408$. Check: $300\le 408\le 800$ ✓ and $340\le 408\le 510$ ✓. A misplaced zero would give $4080$ or $40.8$ — both crash straight through the bounds, so the bounds catch the error automatically.

Apply the distributive law twice. $(t+u)(T+U)=t(T+U)+u(T+U)$ (first: split the height $t+u$ into a top band $t$ and a bottom band $u$). $=tT+tU+uT+uU$ (second: each band's width $T+U$ splits into left column $T$ and right column $U$). Those four terms $tT,tU,uT,uU$ are exactly the four cell areas of the rectangle. So the whole rectangle's area = sum of the four cells — which is precisely what long multiplication adds up. $■$ Picture: two horizontal cuts and two vertical cuts turn one big rectangle into four cells; nothing is added or lost, only re-drawn.

The identity: $9n=10n-n=10(n-1)+(10-n)$. Check the algebra: $10(n-1)+(10-n)=10n-10+10-n=9n$. ✓ Reading the fingers: hold up all $10$ fingers in a row and bend down the $n$-th finger, counting from the left.

• The fingers left of the bent one number $n-1$. The identity assigns them to the tens place, contributing $10\times (n-1)$ — so "count the left fingers, that's how many tens".
• The fingers right of the bent one number $10-n$ (the remaining fingers after removing the $n-1$ on the left and the $1$ bent finger). The identity assigns them to the ones place, contributing $(10-n)$ — so "count the right fingers, that's how many ones".
• Adding the two place-value pieces gives $10(n-1)+(10-n)=9n$, which is exactly $9\times n$. That's why the fingers spell out the answer. Test $9\times 4$: bend finger $4$ → $3$ fingers left, $6$ fingers right → tens digit $3$, ones digit $6$ → $36$. And $9\times 4=36$. ✓

Why legitimate: $25=\frac{100}{4}$, so multiplying by $25$ = multiplying by $100$ then dividing by $4$ (division undoes multiplication, so the ÷4 exactly cancels the extra ×4 hidden in $100=25\times 4$). Compute: $100\times 44=4400$, then $4400÷4=1100$. Independent conceptual check (not just re-arithmetic): $25$ is "a quarter of a hundred", so $25\times 44$ should be one quarter of $44$ hundreds. Split $44$ into four equal parts: each part is $11$, and $11$ hundreds $=1100$. This route never multiplies by $100$ at all — it reasons about quarters — yet lands on the same $1100$, confirming the answer by a genuinely different idea rather than the same sum done twice.

Split each factor by place value, multiply the friendly chunks (using ×10/×9/×5/near-round tricks), then add the chunks back — carrying whenever a column overflows and dropping a placeholder zero whenever a chunk is really "tens". Long multiplication, the area model, and expanding brackets are the same distributive law wearing three costumes. And never forget the trivial anchors: $0$ absorbs (×0 gives 0), $1$ identifies (×1 gives itself).