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

Before we begin, one reminder of the master equation you will use to check everything:

Level 1 — Recognition

(Can you name the parts and read a division statement?)

Q1.

In the statement $53 \div 6 = 8 rem 5$ , name the dividend, divisor, quotient, and remainder.

Recall Solution

Dividend $= 53$ (the whole amount being split).
Divisor $= 6$ (the group size we divide by).
Quotient $= 8$ (how many whole groups).
Remainder $= 5$ (what is left over).

Check: $6 \times 8 + 5 = 48 + 5 = 53$ ✓, and $0 \leq 5 < 6$ ✓.

Q2.

Someone writes $19 \div 4 = 3 rem 7$ . Is this a valid division statement? Explain.

Recall Solution

No. The remainder $7$ is not smaller than the divisor $4$ , so it breaks the rule $0 \leq r < d$ . A leftover of $7$ still contains a full group of $4$ . Bump the quotient up by $1$ and subtract $4$ from the remainder: $3 rem 7 \to 4 rem 3.$ Correct: $19 \div 4 = 4 rem 3$ . Check: $4 \times 4 + 3 = 19$ ✓.

Level 2 — Application

(Do the long division cleanly, column by column.)

Recall the four repeating steps — Divide, Multiply, Subtract, Bring down ("Does McDonald's Sell Burgers?").

Q3.

Compute $847 \div 6$ (quotient and remainder).

Recall Solution

Follow DMSB one place-value column at a time (see figure).

Hundreds: $8 \div 6 = 1$ , $1 \times 6 = 6$ , $8 - 6 = 2$ left. Quotient so far: $1$ .
Bring down the $4$ → running number $24$ . $24 \div 6 = 4$ , $4 \times 6 = 24$ , remainder $0$ . Quotient: $14$ .
Bring down the $7$ → running number $7$ . $7 \div 6 = 1$ , $1 \times 6 = 6$ , $7 - 6 = 1$ left. Quotient: $141$ .
$1 < 6$ , so stop.

Answer: quotient $= 141$ , remainder $= 1$ . Check: $6 \times 141 + 1 = 846 + 1 = 847$ ✓.

Figure — Division — long division, remainder, dividend - divisor - quotient vocabulary

Q4.

Compute $952 \div 7$ .

Recall Solution

Hundreds: $9 \div 7 = 1$ , $9 - 7 = 2$ left.
Bring down $5$ → $25$ . $25 \div 7 = 3$ ( $7 \times 3 = 21 \leq 25 < 28$ ), $25 - 21 = 4$ left.
Bring down $2$ → $42$ . $42 \div 7 = 6$ , remainder $0$ .

Answer: quotient $= 136$ , remainder $= 0$ . Check: $7 \times 136 = 952$ ✓. Remainder $0$ means $7$ divides $952$ exactly.

Level 3 — Analysis

(Watch for placeholder zeros and reversed roles.)

Q5.

Compute $6048 \div 6$ . Be careful.

Recall Solution

Thousands: $6 \div 6 = 1$ , remainder $0$ . Quotient: $1$ .
Bring down $0$ → running number $0$ . $0 \div 6 = 0$ — you must write this $0$ in the quotient. Quotient: $10$ .
Bring down $4$ → running number $4$ . $4 \div 6 = 0$ — again write $0$ (since $4 < 6$ ). Quotient: $100$ .
Bring down $8$ → running number $48$ . $48 \div 6 = 8$ , remainder $0$ . Quotient: $1008$ .

Answer: quotient $= 1008$ , remainder $= 0$ . Check: $6 \times 1008 = 6048$ ✓.

Q6.

Sam wants to split $5$ pizzas among $23$ friends and writes $5 \div 23$ . What are the quotient and remainder, and what does each mean?

Recall Solution

Here the dividend is $5$ and the divisor is $23$ . Since $5 < 23$ , not a single whole group of $23$ fits.

Quotient $= 0$ (each friend gets $0$ whole pizzas).
Remainder $= 5$ (all $5$ pizzas are left over — nobody got a whole one).

Check: $23 \times 0 + 5 = 5$ ✓, and $0 \leq 5 < 23$ ✓.

To actually share them, you'd move to Fractions as Division: each friend gets $\frac{5}{23}$ of a pizza — but as a whole-number division the quotient is $0$ .

Level 4 — Synthesis

(Combine the algorithm with reasoning — build a number to a spec.)

Q7.

Find the smallest number greater than $500$ that leaves remainder $3$ when divided by $8$ .

Recall Solution

"Remainder $3$ when divided by $8$ " means the number has the form $8 q + 3$ (the Division Algorithm run backwards). First find how many $8$ s reach past $500$ : $500 \div 8 = 62 rem 4$ , so $8 \times 62 = 496$ . Try $q = 62$ : $8 \times 62 + 3 = 496 + 3 = 499$ — but $499 < 500$ , too small. Try $q = 63$ : $8 \times 63 + 3 = 504 + 3 = 507$ .

Answer: $507$ . Check: $507 \div 8 = 63 rem 3$ , since $8 \times 63 + 3 = 507$ ✓, and $507 > 500$ ✓.

Q8.

A number $N$ satisfies $N = 9 q + r$ with quotient $q = 12$ and the largest possible remainder. Find $N$ .

Recall Solution

The remainder must satisfy $0 \leq r < 9$ , so the largest possible remainder is $r = 8$ (one less than the divisor). $N = 9 \times 12 + 8 = 108 + 8 = 116.$ Answer: $N = 116$ . Check: $116 \div 9 = 12 rem 8$ , and $8 < 9$ ✓. (If $r$ were $9$ , another group would fit, forcing $q = 13$ .)

Level 5 — Mastery

(Multi-step problems that stress-test every rule at once.)

Q9.

When a number is divided by $6$ the quotient is $14$ and the remainder is $5$ . What remainder does the same number leave when divided by $7$ ?

Recall Solution

First rebuild the number using the Division Algorithm: $N = 6 \times 14 + 5 = 84 + 5 = 89.$ Now divide $89$ by $7$ : $89 \div 7 = 12 rem 5$ , since $7 \times 12 = 84$ and $89 - 84 = 5$ .

Answer: remainder $= 5$ . Checks: $6 \times 14 + 5 = 89$ ✓ and $7 \times 12 + 5 = 89$ ✓, with $5 < 7$ ✓.

Q10.

A three-digit number leaves remainder $2$ when divided by $5$ and remainder $2$ when divided by $3$ . What is the smallest such three-digit number? (Hint: think about what "leaves remainder $2$ under both" says about $N - 2$ .)

Recall Solution

If $N$ leaves remainder $2$ under both $5$ and $3$ , then $N - 2$ is divisible by both $5$ and $3$ — that is, by their LCM, which is $15$ . So $N - 2$ is a multiple of $15$ : $N = 15 k + 2$ . We want the smallest three-digit $N$ , so $N \geq 100$ : $15 k + 2 \geq 100 \Rightarrow 15 k \geq 98 \Rightarrow k \geq 6.53 \dots \Rightarrow k = 7.$ $N = 15 \times 7 + 2 = 105 + 2 = 107.$ Answer: $107$ . Checks: $107 \div 5 = 21 rem 2$ ( $5 \times 21 + 2 = 107$ ) ✓; $107 \div 3 = 35 rem 2$ ( $3 \times 35 + 2 = 107$ ) ✓; and $107$ is three digits ✓.

Q11.

Verify with full long division: does $7$ divide $2 m er g e$ … compute $2401 \div 7$ and state whether $7$ divides it exactly.

Recall Solution

Thousands: $2 \div 7 = 0$ — running number stays, look at $24$ (write $0$ ? No: leading zeros are dropped, so the first real digit uses $24$ ).
$24 \div 7 = 3$ ( $7 \times 3 = 21$ ), $24 - 21 = 3$ left. Quotient: $3$ .
Bring down $0$ → $30$ . $30 \div 7 = 4$ ( $7 \times 4 = 28$ ), $30 - 28 = 2$ left. Quotient: $34$ .
Bring down $1$ → $21$ . $21 \div 7 = 3$ , remainder $0$ . Quotient: $343$ .

Answer: quotient $= 343$ , remainder $= 0$ — so yes, $7$ divides $2401$ exactly. Check: $7 \times 343 = 2401$ ✓. (In fact $2401 = 7^{4}$ , a fact you'd meet in Prime Factorisation.)

Connections

Division — Long Division, Remainder & Vocabulary (parent topic)
Fractions as Division (when the remainder becomes a fraction)
Modular Arithmetic & Remainders (Q9, Q10 in disguise)
HCF and LCM (the LCM trick in Q10)
Prime Factorisation ( $2401 = 7^{4}$ in Q11)
Factors, Multiples & Divisibility Rules
Multiplication — repeated addition (the check step)

Exercises — Division — long division, remainder, dividend - divisor - quotient vocabulary

Level 1 — Recognition

Q1.

Q2.

Level 2 — Application

Q3.

Q4.

Level 3 — Analysis

Q5.

Q6.

Level 4 — Synthesis

Q7.

Q8.

Level 5 — Mastery

Q9.

Q10.

Q11.

Connections

Concept Map