6th Grade · 6.NS.B.2

Multi-digit division in 6th grade

Multi-digit division means dividing large numbers, such as $3{,}192 \div 24$ , with pencil and paper. In 6th grade your child is expected to do long division quickly and correctly, including two-digit divisors, zeros in the answer and remainders.

What the standard says

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

6.NS.B.2 Fluently divide multi-digit numbers using the standard algorithm.

What it means, in plain words

The method most parents learned as long division is called the standard algorithm in school. The number being divided is the dividend, the number you divide by is the divisor and the answer is the quotient. Whatever is left over at the end is the remainder. Long division repeats four steps: divide, multiply, subtract, bring down.

In 4th and 5th grade, children divided with drawings and with partial quotients, a method that takes away easy chunks such as 10 groups at a time. In 6th grade the goal is fluency with multi-digit division: the standard algorithm, done quickly and correctly, including two-digit divisors and long dividends.

Dividing multi-digit numbers well comes down to two things. One is estimating each digit of the quotient, such as how many 24s fit in 79. The other is keeping each digit in its place value. A zero in the middle of the quotient is the most common trap. Remainders matter in word problems, where a remainder can mean a box that is not full or one more bus for the last few students.

Your child should be able to:

Divide a four- or five-digit number by a two-digit number with long division
Estimate each digit of the quotient, such as how many 24s fit in 79
Write a zero in the quotient when the divisor does not fit
Check an answer by multiplying the quotient by the divisor and adding the remainder

Example

What is $3{,}192 \div 24$ ?

24 does not fit into 3, so start with 31. 24 fits into 31 once. Write 1. Then $31 - 24 = 7$ .
Bring down the 9 to make 79. 24 fits 3 times: $3 \times 24 = 72$ and $79 - 72 = 7$ .
Bring down the 2 to make 72. 24 fits 3 times: $72 - 72 = 0$ .
Check: $133 \times 24 = 3{,}192$ . So $3{,}192 \div 24 = 133$ , with no remainder.

Example

A school packs 6,245 pencils into boxes of 31. How many full boxes are there, and how many pencils are left over?

31 fits into 62 twice. Write 2. Then $62 - 62 = 0$ .
Bring down the 4. 31 does not fit into 4, so write 0 in the quotient.
Bring down the 5 to make 45. 31 fits once: $45 - 31 = 14$ .
Check: $201 \times 31 + 14 = 6{,}245$ . There are 201 full boxes and 14 pencils left over.

Where children usually get stuck

For $6{,}245 \div 31$ , your child gets 21 R 14 instead of 201 R 14.
When 31 does not fit into 4, the child brings down the next digit without writing a 0 in the quotient. The digits slide one place to the left, so the answer is ten times too small.
What helps. Estimate before dividing. 31 is close to 30, so the answer should be near $6{,}000 \div 30 = 200$ . An answer of 21 is far from 200, so something went wrong.
For $3{,}192 \div 24$ , your child tries 2 for the 79, subtracts $79 - 48 = 31$ and carries on.
The guess for that digit was too low. The child does not compare what is left over with the divisor, and a leftover of 31 means 24 fits at least once more.
What helps. After each subtraction, ask: is what is left smaller than the divisor? If not, the digit above goes up by one.
Your child's work drifts across the page, and digits end up under the wrong place.
Long division has many rows, and on lined paper the columns are easy to lose. One digit in the wrong column changes the whole answer.
What helps. Have your child do long division on graph paper, one digit per square. The columns stay lined up without any extra effort.

Try four practice questions

Real 6th 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
Look at the long-division work below. One step is missing.
Select the missing step.
A
Quotient digit 3; product 45; difference 3
B
Quotient digit 2; product 30; difference 18
C
Quotient digit 3; product 45; difference 2
D
Quotient digit 4; product 60; difference 12
Show the answer and how to solve it
The missing step is quotient digit $3$ , product $45$ , and difference $3$ .
1. 1The divisor is $15$ and the dividend is $480$ . Since $15$ does not go into the single digit $4$ , the first step uses the first two digits of the dividend: the initial partial dividend is $48$ .
2. 2Compare nearby multiples of $15$ . The greatest one-digit multiple of $15$ that is less than or equal to $48$ is $45$ , which comes from $15 \times 3$ .
3. 3Subtract the product from the partial dividend: $48 - 45 = 3$ .
4. 4So the missing step is quotient digit $3$ , product $45$ , and difference $3$ . Select the option that names this triple.
Common mistake. A common mistake is to look only at the leading digit of the divisor and estimate from there. Here $4 \div 1 \approx 4$ tempts the student to pick $4$ , but $15 \times 4 = 60$ and $60 > 48$ , so $4$ is too large.
The idea underneath. In each long-division step, choose the greatest one-digit multiple of the divisor that is less than or equal to the current partial dividend.
2On level
Look at the division below.
$\displaystyle 523 \div 5$
Select the correct quotient and remainder.
A
$\displaystyle 104\text{ R }3$
B
$\displaystyle 14\text{ R }3$
C
$\displaystyle 140\text{ R }3$
D
$\displaystyle 104\text{ R }1$
Show the answer and how to solve it
The answer is 104 R 3 because 2 is smaller than 5, so a 0 goes in the tens place of the quotient, and 23 divided by 5 leaves a remainder of 3.
1. 1Start with $523 \div 5$ .
2. 25 hundreds divided by 5 is 1 hundred. Next, divide the 2 tens by 5. There are not enough tens to give each group a whole ten, so write 0 in the tens place.
3. 3Bring down the 3 to make 23. 5 goes into 23 four times with 3 left over, since $5 \times 4 = 20$ and $23 - 20 = 3$ .
4. 4So the quotient is 104 R 3. The correct choice is the one that shows 104 R 3.
Common mistake. A common mistake is choosing 14 R 3 and skipping the 0. After 5 goes into 5 once, the next partial dividend is 2, and since 2 is less than 5, you must write 0 in the tens place before bringing down the 3.
The idea underneath. A zero belongs in the quotient whenever the current partial dividend is smaller than the divisor after the quotient has already started.
3On level
Look at the division expression below.
$\displaystyle 43{,}281 \div 6$
Select the correct quotient and remainder.
A
$\displaystyle 7{,}212\text{ R }3$
B
$\displaystyle 7{,}213\text{ R }3$
C
$\displaystyle 7{,}213\text{ R }2$
D
$\displaystyle 7{,}214\text{ R }3$
Show the answer and how to solve it
The answer is 7,213 R 3 because $6 \times 7{,}213$ leaves 3 left over and rebuilds 43,281.
1. 1The dividend is 43,281 and the divisor is 6.
2. 2Multiply the divisor by the quotient 7,213.
3. 3Subtract to find the leftover amount.
4. 4The leftover is 3, and 3 < 6, so 7,213 R 3 is correct. A quotient 1 too low or high shifts the total by 6, and a remainder of 2 is 1 too small.
Common mistake. A common mistake is finding the right quotient 7,213 but recording the wrong remainder after the last subtraction. The remainder must equal 43,281 − 43,278 = 3, and it must be less than 6.
The idea underneath. A quotient-remainder pair is correct only if divisor × quotient + remainder gives the original dividend, with the remainder smaller than the divisor.
4Stretch
Find the quotient and remainder for the division below.
$\displaystyle 2646 \div 13$
Enter the quotient first, then the remainder.
Quotient
Remainder
Show the answer and how to solve it
Enter 203 for the quotient and 7 for the remainder because $13 \times 203 = 2639$ and $2646 - 2639 = 7$ .
1. 1Start with $2646 \div 13$ .
2. 213 goes into 26 exactly two times, so the first quotient digit is 2.
3. 3Subtract 26 and bring down the 4. Because 13 is greater than 4, write 0 in the quotient.
4. 4Bring down the 6 to make 46. Then 13 goes into 46 three times, and 7 is left over.
5. 5So enter 203 as the quotient and 7 as the remainder. Check: $13 \times 203 + 7 = 2639 + 7 = 2646$ . ✓
Common mistake. Leaving out the 0 gives 23 instead of 203. In 203, the 2 means 2 hundreds. In 23, it means 2 tens. Keep the 0 in the tens place.
The idea underneath. A 0 in the quotient is a place holder that keeps the later quotient digits in the correct place value.