4th Grade · 4.NBT.B.6

Long division in 4th grade

Long division is a way to divide a large number, such as 852, by a smaller one in steps. In 4th grade, your child divides numbers with up to four digits by a one-digit number and learns what the leftover amount, called the remainder, means. Your child uses methods that keep the place value in view, such as partial quotients.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.

What it means, in plain words

Division answers two kinds of question: how many are in each group, or how many groups can be made. The number being divided is the dividend, the number you divide by is the divisor and the answer is the quotient. In 4th grade the divisor is always one digit, and the dividend can have up to four digits.

Partial quotients is a method that takes away big, easy chunks first. Your child asks how many 6s fit, starts with a round guess like 100 or 50 and subtracts. This repeats until what is left is smaller than 6, and the chunks are added at the end. Place value helps too: $4{,}200 \div 6$ is 42 hundreds shared by 6, which is 7 hundreds, or 700. Some classes draw the same steps as an area model, a rectangle with the divisor on one side.

Division with remainders comes up when a number does not split evenly. The remainder is what is left over, and it must be smaller than the divisor. In a word problem, your child decides what the remainder means. Sometimes it is left out, sometimes it needs one more box or car and sometimes it is the answer. The standard long division layout, with its divide, multiply, subtract and bring down steps, becomes a goal in 6th grade.

Your child should be able to:

Divide a number with up to four digits by a one-digit number using partial quotients
Use place value to divide large round numbers, such as $3{,}600 \div 9$
Find the remainder and say what it means in a word problem
Check a division by multiplying the quotient by the divisor and adding the remainder

Example

What is $852 \div 6$ ? Use partial quotients.

Start with 100 sixes: $6 \times 100 = 600$ , and $852 - 600 = 252$ .
Next, 40 sixes: $6 \times 40 = 240$ , and $252 - 240 = 12$ .
Last, 2 sixes: $6 \times 2 = 12$ , and $12 - 12 = 0$ .
Add the chunks: $100 + 40 + 2 = 142$ , so $852 \div 6 = 142$ .

Example

175 cookies are packed into boxes that hold 8 cookies each. How many boxes can be filled, and how many cookies are left over?

Fill 20 boxes first: $8 \times 20 = 160$ , and $175 - 160 = 15$ .
One more box: $8 \times 1 = 8$ , and $15 - 8 = 7$ .
7 is less than 8, so it is the remainder: $175 \div 8 = 21 \text{ R } 7$ .
21 boxes can be filled, and 7 cookies are left over. Check: $8 \times 21 + 7 = 175$ .

Where children usually get stuck

For $816 \div 4$ , your child writes 24.
After the 8 hundreds are shared, the 1 ten is too small to share among 4, so the child skips it. That tens place still needs a 0 in the answer, which makes 204.
What helps. Estimate first: 816 is about 800, and $800 \div 4 = 200$ , so 24 is far too small. Then check by multiplying: $4 \times 204 = 816$ .
Your child writes $59 \div 7 = 7 \text{ R } 10$ .
The child stopped one step too early. A remainder of 10 is bigger than 7, so another group of 7 fits. The right answer is 8 R 3.
What helps. After each division, compare the remainder with the divisor. If the remainder is the same or bigger, ask how many more groups fit. Piles of 7 coins make this easy to see.
For 30 children riding in cars that hold 4 each, your child answers "7 R 2 cars" or "7 cars".
The division is right, but the remainder was never read back into the story. The 2 extra children still need a car, so 8 cars are needed.
What helps. After any division word problem, ask your child to answer in a full sentence about the story. Saying "we need 7 cars" out loud makes it clear that 2 children would be left behind.

Try four practice questions

Real 4th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
Use the area model below to divide. Type each side length from left to right, then the quotient and the remainder.
$\displaystyle 84 \div 4$
?
?
4
80
4
Side length 1
Side length 2
Quotient
Remainder
Show the answer and how to solve it
The side lengths are $20$ and $1$ . The quotient is $21$ and the remainder is $0$ .
1. 1Look at the area model. The divisor $4$ is on the left side, and the two rectangles show areas of $80$ and $4$ .
2. 2Divide each rectangle area by the divisor $4$ to find the side length above each rectangle.
3. 3Add the side lengths to find the quotient.
4. 4The rectangles cover $80 + 4 = 84$ . Subtract from the dividend $84$ to find the remainder.
5. 5So $84 \div 4 = 21$ with remainder $0$ .
Common mistake. A common mistake is adding the rectangle areas $80 + 4$ to find the quotient instead of dividing each area by $4$ first. The quotient is the sum of the side lengths, not the sum of the rectangle areas.
The idea underneath. An area model for division splits the dividend into place-value pieces ( $80 + 4 = 84$ ). Dividing each piece by the divisor gives part of the quotient, so the side lengths add up to the whole quotient.
2On level
Liam is solving the division shown below. His quotient and remainder are in the table.
$\displaystyle 683 \div 5$
Quotient Remainder
136 1
Select the statement that best describes Liam's entries.
A
The quotient is correct, but the remainder is incorrect.
B
The quotient is too large by 1.
C
There is no mistake.
D
The remainder is too large; one more group of 5 can be made.
Show the answer and how to solve it
Liam's quotient $136$ is correct, but his remainder $1$ is incorrect. The verification check $5 \times 136 + 1 = 681$ does not equal the dividend $683$ , so the remainder should be $3$ .
1. 1Look at Liam's entries in the table: the quotient is $136$ and the remainder is $1$ .
2. 2First, check whether the remainder is less than the divisor. The remainder $1$ is less than the divisor $5$ , so the remainder-too-large rule does not apply here.
3. 3Now use the verification check: divisor × quotient + remainder should equal the dividend.
4. 4The check gives $681$ , but the dividend is $683$ . Since $5 \times 136 = 680$ , the correct remainder is $683 - 680 = 3$ , not $1$ .
5. 5So Liam's quotient $136$ is correct, but his remainder $1$ is wrong.
Common mistake. A common mistake is to assume that if the remainder is small enough (less than the divisor), the entries must be correct. Liam's remainder $1$ is less than $5$ , but the verification $5 \times 136 + 1 = 681$ does not equal $683$ , so the remainder is still wrong. The remainder size is only one of two checks.
The idea underneath. A correct remainder must be less than the divisor, but that alone is not enough. The full check is: divisor × quotient + remainder = dividend. Use this check whenever you want to confirm a quotient and remainder are right.
3On level
Find the quotient and the remainder. If there is no remainder, enter 0.
$\displaystyle 1{,}600 \div 4$
Quotient Remainder
Show the answer and how to solve it
The quotient is $400$ and the remainder is $0$ , because $1{,}600 \div 4 = 400$ with nothing left over.
1. 1Use place value to think of the dividend. The number 1,600 is the same as 16 hundreds.
2. 2Use the basic fact for 16 divided by 4.
3. 3Because 16 hundreds divided by 4 is 4 hundreds, the quotient is 400.
4. 4Check by multiplying the quotient by the divisor.
5. 5The product equals the dividend exactly, so the remainder is 0. Quotient is 400, remainder is 0.
Common mistake. A common mistake is forgetting the place-value zeros and writing $4$ instead of $400$ . After finding the basic fact $16 \div 4 = 4$ , keep the same number of trailing zeros from the dividend in the quotient.
The idea underneath. Dividing a zero-ending number by a one-digit divisor uses the same basic fact as dividing the leading digits, but the place-value zeros stay in the quotient. Multiplying the quotient by the divisor must give the dividend back exactly.
4Stretch
A student drew the array below to show $63 \div 7$ .
The student writes that $63 \div 7 = 7$ with a remainder of $0$ .
Select the statement that best describes the student's work.
A
The model uses the wrong number of rows for the divisor.
B
The quotient was counted from the rows instead of the columns.
C
The leftover count was not handled correctly.
D
The division is correct.
Show the answer and how to solve it
The array has $9$ rows but the divisor is $7$ , so the row count does not match the divisor. The correct result is $63 \div 7 = 9$ with a remainder of $0$ .
1. 1Find the correct quotient and remainder for $63 \div 7$ .
2. 2Look at the student's array. It has $9$ rows and $7$ columns with $0$ extra tiles. The total is $9 \times 7 + 0 = 63$ , which matches the dividend. However, the divisor is $7$ , so the array should have $7$ rows — not $9$ rows.
3. 3The student claims $63 \div 7 = 7$ with a remainder of $0$ . The correct remainder is $0$ , but the correct quotient is $9$ . The array has $9$ rows (the actual quotient) and $7$ columns (the divisor) — the rows and columns are swapped. Because rows ≠ divisor, the array structure is invalid.
4. 4Because the array has $9$ rows instead of $7$ rows for the divisor, the best description is: 'The model uses the wrong number of rows for the divisor.'
Common mistake. A common mistake is putting the divisor in the columns instead of the rows. In a division array, rows = divisor.
The idea underneath. In a division array, rows show the divisor, columns show the quotient, and any leftover tiles show the remainder.

Side length 1
Side length 2
Quotient
Remainder