Division — long division, remainder, dividend - divisor - quotient vocabulary: Worked examples

The scenario matrix

Before working anything, let us name every distinct shape a division problem can take. Each row is a "cell" — a genuinely different situation. If we cover all cells, no exam question is a stranger.

#	Cell (case class)	What makes it different	Example we use
A	Clean divide, remainder $0$	Nothing left over; divisor divides dividend	$945 \div 7$ (revisited deeper)
B	Leftover at every column	A non-zero remainder marches through each step	$853 \div 6$
C	Placeholder zero inside the quotient	Running number smaller than divisor mid-way → must write $0$	$6048 \div 6$
D	Divisor bigger than the leading digit(s)	Quotient "starts late"	$284 \div 7$
E	Divisor bigger than the WHOLE dividend	Quotient is $0$ , dividend is its own remainder	$5 \div 8$
F	Dividing zero	$0$ shared into groups → everything zero	$0 \div 4$
G	Divide by $1$ / by the number itself	Degenerate divisors — sanity anchors	$37 \div 1$ , $37 \div 37$
H	Two-digit divisor (exam twist)	Estimation matters more; DMSB unchanged	$4732 \div 23$
I	Real-world word problem	You must decide what quotient and remainder mean	Buses & passengers

We will hit A–I, in that order. Watch the running condition $0 \leq Remainder < Divisor$ hold in every single one — that is the thread tying them together.

Cell A — Clean divide (remainder $0$ )

Worked example

$945 \div 7$ Forecast: Guess before dividing. $7 \times 100 = 700$ , $7 \times 130 = 910$ , $7 \times 135 = 945$ ? We suspect quotient is around $135$ and — because $945$ feels like it might be a multiple of $7$ — maybe remainder $0$ . Hold that guess.

Step 1 — Divide the hundreds. $9 \div 7 = 1$ , write $1$ . Then $1 \times 7 = 7$ , subtract $9 - 7 = 2$ . Why this step? We ask "how many whole $7$ s fit in $9$ hundreds?" — exactly one hundred-group, with $2$ hundreds left to carry down.

Step 2 — Bring down the tens. The leftover $2$ hundreds $= 20$ tens; the number had $4$ tens, giving $24$ . Now $24 \div 7 = 3$ (since $7 \times 3 = 21 \leq 24 < 28$ ), write $3$ , subtract $24 - 21 = 3$ . Why this step? "Bring down" is just place value: a leftover hundred is ten tens, so it joins the tens column.

Step 3 — Bring down the ones. Leftover $3$ tens $= 30$ ones, plus the $5$ ones $= 35$ . Then $35 \div 7 = 5$ exactly, subtract $35 - 35 = 0$ . Why this step? Same reasoning on the ones. Remainder $0 < 7$ — we stop, cleanly.

Answer: quotient $= 135$ , remainder $= 0$ . Verify: $7 \times 135 + 0 = 945$ ✓. Remainder $0$ means $7$ ==divides $945$ exactly== — see Factors, Multiples & Divisibility Rules.

Cell B — Leftover marching through every column

Worked example

$853 \div 6$ Forecast: $6 \times 100 = 600$ , $6 \times 140 = 840$ , so quotient is in the $140$ s; and $853$ is not obviously a multiple of $6$ , so expect a non-zero remainder. Guess ~ $142$ rem something.

Look at the figure above: each column keeps a red leftover that gets carried right. Nothing ever lands on $0$ until the end.

Step 1 — Hundreds. $8 \div 6 = 1$ , $1 \times 6 = 6$ , subtract $8 - 6 = 2$ (the first red leftover). Why? One $6$ fits in $8$ hundreds, $2$ hundreds remain.

Step 2 — Tens. Bring down $5$ : leftover $2$ hundreds $= 20$ tens, plus $5$ tens $= 25$ . $25 \div 6 = 4$ ( $6 \times 4 = 24 \leq 25 < 30$ ), subtract $25 - 24 = 1$ (red leftover again). Why? Place value carries the leftover into the next column.

Step 3 — Ones. Bring down $3$ : $10 + 3 = 13$ . $13 \div 6 = 2$ ( $6 \times 2 = 12 \leq 13 < 18$ ), subtract $13 - 12 = 1$ . Why? $1 < 6$ , so we stop — the condition $0 \leq r < 6$ is satisfied.

Answer: quotient $= 142$ , remainder $= 1$ . Verify: $6 \times 142 + 1 = 852 + 1 = 853$ ✓. Forecast ( $142$ ) nailed.

Cell C — The placeholder zero (the trap)

Worked example

$6048 \div 6$ Forecast: $6 \times 1000 = 6000$ , so the quotient is just over $1000$ . This matters: if your answer comes out three digits long, you dropped a zero.

Step 1 — Thousands. $6 \div 6 = 1$ , write $1$ , subtract $6 - 6 = 0$ .

Step 2 — Hundreds (THE DANGER). Bring down $0$ : running number is $0$ . Now $0 \div 6 = 0$ — you must write a $0$ in the quotient. Why this step? If you skip writing that $0$ and jump ahead, the digits shift and $1008$ collapses into $18$ . The $0$ is a genuine place-holder saying "zero hundred-groups here."

Step 3 — Tens. Bring down $4$ : running number $04 = 4$ . Since $4 < 6$ , again $4 \div 6 = 0$ — write another $0$ , leftover $4$ . Why? Same rule: bring down a digit, if the running number is smaller than the divisor, the quotient digit is $0$ .

Step 4 — Ones. Bring down $8$ : leftover $4$ tens $= 40$ ones, plus $8 = 48$ . $48 \div 6 = 8$ exactly, subtract $0$ .

Answer: quotient $= 1008$ , remainder $= 0$ . Verify: $6 \times 1008 = 6048$ ✓. Count the digits: $1, 0, 0, 8$ — four digits, matching our forecast of "just over $1000$ ."

Cell D — Divisor bigger than the leading digit

Worked example

$284 \div 7$ Forecast: $7 \times 40 = 280$ , so quotient is around $40$ . Note it will be two digits, not three, even though the dividend has three digits.

Step 1 — Hundreds. $2 \div 7 = 0$ ? Seven can't fit in $2$ . So there are zero hundred-groups; we don't write a leading $0$ (leading zeros are dropped), we simply look at $28$ instead. Why this step? The very first quotient digit is only written once we have enough to make a group. $2$ hundreds isn't enough, so the quotient starts at the tens place.

Step 2 — Look at $28$ (tens). $28 \div 7 = 4$ exactly, subtract $28 - 28 = 0$ . Why? $7 \times 4 = 28$ , a perfect fit.

Step 3 — Ones. Bring down $4$ : running number $4$ . $4 \div 7 = 0$ , write $0$ , leftover $4$ . Why? $4 < 7$ so no group forms — but this $0$ is inside the quotient (a real digit), not a dropped leading zero.

Answer: quotient $= 40$ , remainder $= 4$ . Verify: $7 \times 40 + 4 = 280 + 4 = 284$ ✓.

Cell E — Divisor bigger than the WHOLE dividend

Cell F — Dividing zero

Cell G — Degenerate divisors ( $1$ and the number itself)

Cell H — Two-digit divisor (the exam twist)

Worked example

$4732 \div 23$ Forecast: Estimation now does real work. $23 \approx 20$ , and $4732/20 \approx 236$ . So quotient is near $205$ – $210$ ? Refine: $23 \times 200 = 4600$ , leaving $132$ ; $23 \times 5 = 115$ ; so ~ $205$ with a small remainder. Let's confirm.

The same DMSB loop from the parent note (Divide, Multiply, Subtract, Bring down) runs — only the "Divide" sub-step needs a guess-and-check because $23$ isn't a single digit. The red boxes in the figure mark where we estimate.

Step 1 — Take the first two digits, $47$ . How many $23$ s in $47$ ? Estimate $47 \div 23 \approx 2$ . Check: $23 \times 2 = 46 \leq 47$ , and $23 \times 3 = 69 > 47$ . So digit $= 2$ . Subtract $47 - 46 = 1$ . Why estimate? With a two-digit divisor you can't just "know" the digit; you round ( $23 \to 20$ , $47 \to 46$ ) and verify it fits.

Step 2 — Bring down $3$ : running number $= 13$ . How many $23$ s in $13$ ? None. Write $0$ , leftover $13$ . Why? $13 < 23$ , so this quotient digit is a genuine placeholder $0$ (Cell C behaviour reappearing).

Step 3 — Bring down $2$ : running number $= 132$ . Estimate $132 \div 23$ : $23 \times 5 = 115 \leq 132$ , $23 \times 6 = 138 > 132$ . Digit $= 5$ . Subtract $132 - 115 = 17$ . Why? Round $23 \to 20$ : $132/20 \approx 6$ , try $6$ — too big — step back to $5$ . This "guess, over-shoot, step back" is normal for multi-digit divisors.

Answer: quotient $= 205$ , remainder $= 17$ . Verify: $23 \times 205 + 17 = 4715 + 17 = 4732$ ✓, and $0 \leq 17 < 23$ ✓. Forecast (~ $205$ ) landed exactly.

Cell I — Real-world word problem (deciding what the remainder means)

Worked example A school has

$329$ students. Each bus seats $40$ . How many buses are needed? Forecast: $40 \times 8 = 320$ , so about $8$ buses — but $9$ students are still standing. Do we round up or drop them? Think before dividing.

Step 1 — Divide. $329 \div 40$ : how many $40$ s in $329$ ? $40 \times 8 = 320 \leq 329$ , $40 \times 9 = 360 > 329$ . Quotient $= 8$ , remainder $329 - 320 = 9$ . Why? Standard division: $8$ full buses carry $320$ students.

Step 2 — Interpret the remainder. $9$ students remain with no seat. You cannot leave them behind, so you need one more bus. Why this step? The maths gives $q = 8$ , $r = 9$ — but the word problem asks "how many buses," and a partial group of students still needs a whole bus. So the answer is $8 + 1 = 9$ buses.

Answer: $9$ buses (the last carries only $9$ students). Verify: $8$ buses $\times 40 = 320$ seats, plus $1$ bus for the last $9 \Rightarrow$ capacity $360 \geq 329$ ✓; with $8$ buses only $320 < 329$ ✗. So $9$ is genuinely the minimum.

Recall checkpoint

Recall Which cell is each of these?

$6048 \div 6 = 1008$ — which trap does it teach? ::: Cell C — the placeholder zero (skip it and you get $18$ ). Why is $5 \div 8$ quotient $0$ remainder $5$ ? ::: Cell E — divisor exceeds dividend, so no whole group fits; the whole dividend is the remainder. Is $4 \div 0$ the same trap as $0 \div 4$ ? ::: No — $0 \div 4 = 0$ is fine; $4 \div 0$ is undefined (nothing times $0$ gives $4$ ). In the bus problem, why answer $9$ not $8$ ? ::: The remainder $9$ students still need a bus, so round the quotient up. What single check verifies every example above? ::: $Divisor \times Quotient + Remainder = Dividend$ , with $0 \leq r < divisor$ .

Connections

Parent: Division & vocabulary
Fractions as Division (Cell E: a leftover becomes a fraction)
Modular Arithmetic & Remainders (the remainder, studied on its own)
Factors, Multiples & Divisibility Rules (remainder $0 \Rightarrow$ divides exactly)
Multiplication — repeated addition (the verify step is multiplication)

Worked examples — Division — long division, remainder, dividend - divisor - quotient vocabulary

The scenario matrix

Cell A — Clean divide (remainder $0$ )

Cell B — Leftover marching through every column

Cell C — The placeholder zero (the trap)

Cell D — Divisor bigger than the leading digit

Cell E — Divisor bigger than the WHOLE dividend

Cell F — Dividing zero

Cell G — Degenerate divisors ( $1$ and the number itself)

Cell H — Two-digit divisor (the exam twist)

Cell I — Real-world word problem (deciding what the remainder means)

Recall checkpoint

Connections

Case Map