5th Grade · 5.NBT.B.6

Long division with 2-digit divisors in 5th grade

Long division with 2-digit divisors means dividing by a number from 10 to 99, such as $864 \div 24$ . In 5th grade, your child divides numbers of up to four digits by two-digit numbers. The standards ask for methods built on place value, such as partial quotients and area models. The fast written method many parents learned is a 6th grade goal.

What the standard says

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

5.NBT.B.6 Find whole-number quotients of whole numbers with up to four-digit dividends and two-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 has its own words. The dividend is the number being divided, the divisor is the number you divide by and the quotient is the answer. The remainder is what is left over when the divisor does not fit exactly. In $864 \div 24 = 36$ , 864 is the dividend, 24 is the divisor and 36 is the quotient.

Division with two-digit divisors is harder than dividing by one digit. The times tables no longer give the answer directly, so your child has to estimate. With partial quotients, your child takes away easy chunks, such as 10, 20 or 5 times the divisor. This goes on until what is left is smaller than the divisor. Adding up the chunks gives the quotient.

Long division with 2-digit divisors ends with a check by multiplication, because multiplying undoes dividing. The divisor times the quotient, plus the remainder, must give back the dividend. An area model shows the same idea as a rectangle. The dividend is the area, the divisor is one side and the quotient is the other side.

Your child should be able to:

Divide a number of up to four digits by a two-digit number
Use partial quotients or an area model and explain each step
Estimate the quotient first by rounding, such as treating 31 as 30
Check an answer: the divisor times the quotient, plus the remainder, gives the dividend

Example

What is $864 \div 24$ ?

Take an easy chunk first: $24 \times 30 = 720$ .
Subtract: $864 - 720 = 144$ . That is still more than 24, so keep going.
$24 \times 6 = 144$ . Subtract: $144 - 144 = 0$ .
Add the chunks: $30 + 6 = 36$ . Check: $24 \times 36 = 864$ .
So 864 divided by 24 is 36.

Example

A school packs 1,550 crayons into boxes of 32. How many boxes can it fill, and how many crayons are left over?

The problem is $1{,}550 \div 32$ .
Take a big chunk: $32 \times 40 = 1{,}280$ . Subtract: $1{,}550 - 1{,}280 = 270$ .
Take another chunk: $32 \times 8 = 256$ . Subtract: $270 - 256 = 14$ .
14 is less than 32, so stop. The quotient is $40 + 8 = 48$ with a remainder of 14.
Check: $32 \times 48 + 14 = 1{,}536 + 14 = 1{,}550$ .
The school fills 48 boxes and has 14 crayons left over.

Where children usually get stuck

Your child answers $437 \div 24$ with 17 R 29.
The child stopped taking away chunks too early. A remainder of 29 means one more 24 still fits, so the answer is 18 R 5.
What helps. After every division, ask whether the remainder is smaller than the divisor. If it is not, one more group fits.
Your child erases guess after guess, trying $24 \times 7$ , then $24 \times 9$ , then $24 \times 8$ .
With a two-digit divisor, the times tables no longer give the next digit. Guessing without a plan wears the child out.
What helps. Before starting, write a short list of easy multiples of the divisor: $24 \times 10 = 240$ , $24 \times 20 = 480$ , $24 \times 5 = 120$ . Your child then picks chunks from the list.
Your child divides 2,460 by 12 and writes 25 instead of 205.
In the written method, 12 does not fit into 6 tens, so the tens digit of the quotient is 0. The child skips that step and the 0 goes missing.
What helps. Estimate first: $12 \times 200 = 2{,}400$ , so the answer must be a bit more than 200. An answer of 25 is clearly too small.

Try four practice questions

Real 5th 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

Jake wants to decide whether 39 R 18 makes sense as an answer for

840 \div 21

. His check is shown below. An answer makes sense here when the quotient is close to the correct estimate and the remainder is less than the divisor.

Round	$\displaystyle 840 \div 21 \approx 800 \div 20$
Estimate	$\displaystyle \approx 40$
Conclusion	$\displaystyle \text{So } 39 \text{ R } 18 \text{ is reasonable.}$

Select the statement that best describes Jake's work.

The student made a rounding mistake because 840 should have been rounded to 1,000 instead of 800.

The student's work is correct.

The student rounded correctly, but

800 \div 20

is 40, so 39 R 18 is not reasonable.

The student's conclusion is wrong because 39 R 18 is not the exact answer to

840 \div 21

Show the answer and how to solve it

The student's work is correct. The rounding, estimate, and conclusion all hold up.

1Jake's rounding is sensible. Changing $840 \div 21$ to $800 \div 20$ uses nearby compatible numbers.
2Verify the estimate from the rounded numbers.
3Compare the proposed answer to the estimate. The quotient 39 is only 1 away from 40, and the remainder 18 is less than the divisor 21.
4The rounding, estimate, and conclusion are all correct, so Jake's work is correct.

Common mistake. A common mistake is assuming the work must contain an error. Here the rounding to $800 \div 20$ , the estimate 40, and the reasonable conclusion all hold up.

The idea underneath. An answer makes sense when the quotient is close to the correct estimate and the remainder is smaller than the divisor. If a remainder is not smaller, you can make another full group.

2On level
Alex is solving $638 \div 18$ using partial quotients. His work is shown below.
638
− 540 18 × 30
98
− 90 18 × 5
12
remainder
35
quotient
Select the statement that best describes Alex's work.
A
The student's work is correct.
B
The final quotient total is incorrect.
C
The first partial-quotient step is incorrect.
D
The second partial-quotient step is incorrect.
Show the answer and how to solve it
The correct choice is "The second partial-quotient step is incorrect." The first row and the total 35 are correct, but $98 - 90$ should be 8, not 12.
1. 1Check each row from top to bottom before deciding.
2. 2The first step is correct.
3. 3The second product is correct, but the second subtraction should give 8, not 12.
4. 4The partial quotients 30 and 5 count groups of 18. Adding the group counts gives 35 groups. Adding the partial quotients gives the correct total.
Common mistake. A common mistake is checking only the products $18 \times 30$ and $18 \times 5$ . In partial quotients, each new remainder must also come from subtraction, so $98 - 90$ has to be checked too.
The idea underneath. Each partial quotient counts groups of the divisor. Add these group counts for the quotient, and check each multiplication and subtraction.
3On level
A student is dividing with partial quotients. In the work below, explain the second subtraction step.
252
− 150 15 × 10
102
− 90 15 × 6
12
remainder
16
quotient
Use the dropdowns to complete the explanation below.
The number being subtracted in the second step is:
This is true because the matching equation is:
Show the answer and how to solve it
The number being subtracted in the second step is a multiple of the divisor because $15 \times 6 = 90$ .
1. 1Find the second subtraction row in the workspace. After the first step, 102 is left; the second row subtracts 90 and leaves 12.
2. 2Use the divisor and the partial quotient from the second row.
3. 3Since 90 is made by multiplying the divisor 15 by 6, the number 90 is a multiple of the divisor.
Common mistake. A common mistake is calling 90 the remainder because it appears in the subtraction area. The remainders around the second step are 102 before and 12 after, while 90 is the amount subtracted.
The idea underneath. Each subtraction in partial quotients removes a chunk equal to the divisor multiplied by that row's partial quotient.
4Stretch
The completed work below shows $672 \div 23$ with partial quotients.
672
− 460 23 × 20
212
− 161 23 × 7
51
− 46 23 × 2
5
remainder
29
quotient
Fill in the blanks to match the work, in step order. Each step appears as a multiplication or its product. Add the remainder only when it is not zero.
$\displaystyle 672$
$\displaystyle =$
$\displaystyle 23 \times$
$\displaystyle + 161 +$
$\displaystyle +$
Show the answer and how to solve it
The blanks are $20$ , $46$ , and $5$ , so the equation is $672 = 23 \times 20 + 161 + 46 + 5$ .
1. 1Read the completed work. The divisor is $23$ , the partial quotients are $20$ , $7$ , and $2$ , and the remainder is $5$ .
2. 2Multiply each partial quotient by the divisor to find the amount subtracted.
3. 3The equation writes the first chunk as $23 \times$ blank, the second chunk's product $161$ is already shown, the third chunk's product $46$ goes in the second blank, and the remainder $5$ goes in the third blank.
4. 4So the blanks, in order, are $20$ , $46$ , and $5$ .
Common mistake. A common mistake is writing the total quotient $29$ in the first blank instead of the partial quotient $20$ . Another mistake is writing $2$ (the third partial quotient) in the second blank instead of the product $46$ .
The idea underneath. Partial-quotients work already breaks the dividend into divisor multiples and a remainder, so the equation is just a one-line record of those same chunks.

	672
−	460	23	×	20

	212
−	161	23	×	7

	51
−	46	23	×	2

	5 remainder			29 quotient