Primary & Middle School Mathematics · Grades 1–9

23Division

Grade 3 divided with the tables and small remainders (Chapter 17); this chapter installs the full written method — long division — which handles any number, digit group by digit group.

23.1 The long division method

Method 23.1 (Long division by a one-digit number)

To divide $749$ by $6$ :

take the leftmost digit, $7$ (hundreds): $6$ goes $1$ time into $7$ ; write $1$ in the quotient, subtract $6$ , remainder $1$ ;
bring down the next digit, $4$ : now divide $14$ (tens): $6 \times 2 = 12$ ; write $2$ , remainder $2$ ;
bring down the $9$ : divide $29$ : $6 \times 4 = 24$ ; write $4$ , remainder $5$ ;
no digit left: the quotient is $124$ , the remainder $5$ .

Check: $6 \times 124 + 5 = 744 + 5 = 749$ , and $5 < 6$ .

The long division of 749 by 6, written in full: each subtraction appears, every digit stays in its column, and the gray arrows show the digits being brought down. — The long division of $749$ by $6$ , written in full: each subtraction appears, every digit stays in its column, and the gray arrows show the digits being brought down.

Example 23.2 (A zero in the quotient)

Divide $618$ by $3$ : hundreds, $6 \div 3 = 2$ ; tens, bring down the $1$ : $3$ goes $0$ times into $1$ — write the $0$ ! — remainder $1$ ; units, bring down the $8$ : $18 \div 3 = 6$ . Quotient $206$ , remainder $0$ . Forgetting the middle zero (writing $26$ ) is the classic mistake; the check $3 \times 26 = 78 \neq 618$ catches it immediately.

Example 23.3 (Estimating the quotient’s size)

Before dividing $749$ by $6$ , frame the answer: $6 \times 100 = 600$ and $6 \times 200 = 1\,200$ , so the quotient is between $100$ and $200$ — it will have three digits. This one-line estimate prevents most misplaced-digit errors.

23.2 Choosing what the question asks

Method 23.4 (Quotient, remainder, or quotient plus one)

As in Method 17.5, the story decides:

“how many full …” or “how many each”: the quotient;
“how many left”: the remainder;
“how many containers needed for all”: the quotient, plus one if the remainder is not zero.

Example 23.5

$530$ books must be packed in boxes of $8$ .

530 = 8 \times 66 + 2 .

How many full boxes? $66$ . How many books are not in a full box? $2$ . How many boxes to pack all the books? $67$ . Three questions, one division.

Example 23.6 (Fair sharing of money)

Four friends share the cost of a $92$ gift equally: $92 \div 4 = 23$ exactly (remainder $0$ ). Each pays $23$ . When the remainder is zero, the division “comes out even” and the sharing is perfectly fair.

23.3 Exercises

Exercise 23.1 ★

Frame each quotient as in Example 23.3 (between which two “round” numbers?), then compute the long division: $96 \div 4$ ; $87 \div 5$ .

Solution

Solution of Exercise 23.1.

$96 \div 4$ : between $4 \times 20 = 80$ and $4 \times 30 = 120$ , so the quotient has two digits; long division gives $24$ , remainder $0$ .

$87 \div 5$ : between $5 \times 10 = 50$ and $5 \times 20 = 100$ ; quotient $17$ , remainder $2$ ( $87 = 5 \times 17 + 2$ ).

Exercise 23.2 ★

Compute the long divisions and check each one: $672 \div 4$ ; $925 \div 7$ ; $804 \div 6$ .

Solution

Solution of Exercise 23.2.

$672 \div 4 = 168$ , remainder $0$ (check: $4 \times 168 = 672$ ).

$925 \div 7 = 132$ , remainder $1$ (check: $7 \times 132 + 1 = 925$ ).

$804 \div 6 = 134$ , remainder $0$ (check: $6 \times 134 = 804$ ).

Exercise 23.3 ★

Careful with the zero (see Example 23.2): $816 \div 4$ ; $420 \div 7$ ; $2\,512 \div 5$ .

Solution

Solution of Exercise 23.3.

$816 \div 4 = 204$ (the tens step is $1 \div 4$ : zero times — write the $0$ ).

$420 \div 7 = 60$ .

$2\,512 \div 5 = 502$ , remainder $2$ (check: $5 \times 502 + 2 = 2\,512$ ).

Exercise 23.4 ★

Write the check equality ( $a = b \times q + r$ ) for: $85 \div 9$ ; $1\,000 \div 3$ .

Solution

Solution of Exercise 23.4.

$85 = 9 \times 9 + 4$ (quotient $9$ , remainder $4$ ).

$1\,000 = 3 \times 333 + 1$ .

Exercise 23.5 ★

$156$ eggs are put in boxes of $6$ . How many boxes are filled?

Solution

Solution of Exercise 23.5.

$156 \div 6 = 26$ : twenty-six boxes are filled, none left over.

Exercise 23.6 ★

A ribbon of $250$ cm is cut into pieces of $8$ cm. How many pieces, and what length is left over?

Solution

Solution of Exercise 23.6.

$250 = 8 \times 31 + 2$ : thirty-one pieces, and $2$ cm of ribbon left.

Exercise 23.7 ★

$375$ students go to a show in buses of $52$ seats. How many buses are needed? (Which case of Method 23.4 is this? Divide by $52$ using multiples: $52 \times 7 = 364$ .)

Solution

Solution of Exercise 23.7.

$375 = 52 \times 7 + 11$ : seven buses carry $364$ students, $11$ remain — the “plus one” case: $8$ buses are needed.

Exercise 23.8 ★

Three friends share $234$ marbles equally. How many marbles each? Check with a multiplication.

Solution

Solution of Exercise 23.8.

$234 \div 3 = 78$ : each gets $78$ marbles. Check: $3 \times 78 = 234$ .

Exercise 23.9 ★

Find the dividend: the division by $7$ gives quotient $58$ and remainder $3$ .

Solution

Solution of Exercise 23.9.

Dividend $= 7 \times 58 + 3 = 406 + 3 = 409$ .

Exercise 23.10 ★★

A librarian shelves $438$ books, $9$ per shelf. Shelves come in bookcases of $6$ shelves. How many shelves are needed? How many bookcases must be bought? (Two divisions, both with a “plus one” question.)

Solution

Solution of Exercise 23.10.

Shelves: $438 = 9 \times 48 + 6$ : $48$ full shelves and $6$ books more — $49$ shelves are needed. Bookcases: $49 = 6 \times 8 + 1$ : eight full bookcases and one extra shelf — $9$ bookcases must be bought.

Exercise 23.11 ★★

In a division by $8$ , the quotient equals the remainder. What are the possible dividends? List them all. (The remainder must stay below $8$ .)

Solution

Solution of Exercise 23.11.

Quotient $=$ remainder $= r$ with $r < 8$ , so $r$ is $0, 1, \dots, 7$ and the dividend is $8r + r = 9r$ : the possible dividends are $0$ , $9$ , $18$ , $27$ , $36$ , $45$ , $54$ , $63$ .