Divide by 2-Digit Numbers One Place at a Time | Grade 5 Math | LevelKite

You can already divide by a 2-digit number with partial quotients. You take away chunks and add up the partial quotients at the end. Now you will learn a shorter way to write the same work. You divide one place at a time and write the answer one digit at a time.

First, here are four words you need. Look at $87 \div 21$ .

The dividend is $87$ , the number being divided.
The divisor is $21$ , the number you divide by.
The quotient is the answer. Picture $87$ shared equally into $21$ groups. The quotient is how many each group gets.
The remainder is what is left over when no more can be shared equally.

Each group can get $4$ , because $4 \times 21 = 84$ . They cannot get $5$ each, because $5 \times 21 = 105$ is more than $87$ . So the quotient is $4$ , and $87 - 84 = 3$ is left over. You write $87 \div 21 = 4$ R $3$ . The R stands for remainder.

Here the quotient has just one digit, in the ones place. A bigger dividend can give a quotient with more digits.

Before you divide, decide where the first digit of the quotient goes. Place value tells you.

Take $745 \div 32$ . The number $745$ is 7 hundreds, 4 tens and 5 ones. Share it into $32$ equal groups.

There are only 7 hundreds for 32 groups. That is not enough to give every group a whole hundred, so the quotient has no hundreds digit.
Trade the 7 hundreds for 70 tens. With the 4 tens, that makes 74 tens. Now every group can get some tens. The first digit goes in the tens place.

Here is a quick way to decide. Read the first two digits of the dividend as one number.

If it is equal to or more than the divisor, divide those two digits first.
If it is less than the divisor, there is not enough yet. Trade once more and divide the first three digits. In $245 \div 32$ there are only 24 tens for 32 groups, so you divide 245 ones.

The first digit of the quotient goes in the place of the last digit you used. This quick way works when the divisor has 2 digits and the dividend is at least as big as the divisor.

Key concept

One place at a time, one digit for each place

Think of dividing as sharing into equal groups, one place at a time. Start at the place of the first quotient digit. Then repeat these steps for each place, down to the ones.

Divide. Find the most each group can get. It is the biggest digit you can multiply by the divisor without going over what you have.
Multiply that digit by the divisor. This is how much you used.
Subtract to see what is left. It must be less than the divisor. If it were not, each group could get one more.
Trade what is left for the next smaller place, and join it with the next digit of the dividend. This is called bringing down the digit.

Each of these places gets exactly one digit of the quotient. At the ones place there is no trade: what is left is the remainder.

Worked example

A 3-digit dividend

Find $745 \div 32$ .

Decide where the first digit of the quotient goes. The first two digits make 74, and 74 is more than 32. So divide the 74 tens first. The first digit goes in the tens place, and the quotient will have 2 digits.
$74 > 32$
Divide the 74 tens. Each group can get 2 tens, because 2 × 32 = 64. They cannot get 3 tens each, because 3 × 32 = 96 is more than 74. Write 2 in the tens place. Subtract the 64 tens you used. This is the same as taking away 20 × 32 = 640 with partial quotients. The 10 tens left are fewer than 32, so move on.
$74 \text{ tens} - 64 \text{ tens} = 10 \text{ tens}$
Trade the 10 tens for 100 ones and join them with the 5 ones. Now you have 105 ones. Each group can get 3 ones, because 3 × 32 = 96. They cannot get 4 each, because 4 × 32 = 128 is more than 105. Write 3 in the ones place and subtract the 96 ones you used.
$105 \text{ ones} - 96 \text{ ones} = 9 \text{ ones}$
The ones place is the last place, so the 9 left over is the remainder. It is less than 32, so there is not enough to give every group one more. Read the quotient: 2 tens and 3 ones make 23.
$745 \div 32 = 23 \text{ R } 9$

Answer

$745 \div 32 = 23$ R $9$ . Each of the $32$ groups gets $23$ , and $9$ are left over.

Where the first quotient digit goes. The last two rows share a divisor but start in different places.

Division	First two digits and the divisor	Divide first	First quotient digit goes in the	Answer
$87 \div 21$	$87$ is more than $21$	87 ones	ones place	$4$ R $3$
$745 \div 32$	$74$ is more than $32$	74 tens	tens place	$23$ R $9$
$245 \div 32$	$24$ is less than $32$	245 ones	ones place	$7$ R $21$ : $7 \times 32 = 224$ fits, $8 \times 32 = 256$ is too much, and $245 - 224 = 21$

A 4-digit dividend works the same way. It has more places, so you repeat the steps more times.

You still decide first where the first digit of the quotient goes. When you divide a 4-digit number by a 2-digit number, there are two cases.

The first two digits count hundreds. If they make a number equal to or more than the divisor, the first digit goes in the hundreds place. The quotient has 3 digits.
If they make a number less than the divisor, use the first three digits, which count tens. The first digit goes in the tens place. The quotient has 2 digits.

One new thing can happen along the way. What you have at a place may be less than the divisor. Then there is not enough to give every group one of that place, so the digit for that place is $0$ . Once the quotient has its first digit, you must write this $0$ . It holds the place so the other digits keep their values.

Worked example

A 4-digit dividend with a 0 in the quotient

Find $5{,}347 \div 26$ .

Decide where the first digit of the quotient goes. The first two digits make 53, and 53 is more than 26. So divide the 53 hundreds first. The first digit goes in the hundreds place. Each group can get 2 hundreds, because 2 × 26 = 52. They cannot get 3 each, because 3 × 26 = 78 is more than 53. Write 2 in the hundreds place and subtract.
$53 \text{ hundreds} - 52 \text{ hundreds} = 1 \text{ hundred}$
Trade the 1 hundred for 10 tens and join them with the 4 tens. Now you have 14 tens. 14 is less than 26, so there are not enough tens to give every group one. Write 0 in the tens place. The same steps still work: 0 × 26 = 0, so all 14 tens are left.
$14 \text{ tens} - 0 \text{ tens} = 14 \text{ tens}$
Trade the 14 tens for 140 ones and join them with the 7 ones. Now you have 147 ones. Each group can get 5 ones, because 5 × 26 = 130. They cannot get 6 each, because 6 × 26 = 156 is more than 147. Write 5 in the ones place and subtract.
$147 \text{ ones} - 130 \text{ ones} = 17 \text{ ones}$
The ones place is the last place, so 17 is the remainder. It is less than 26. Read the quotient in order: 2 hundreds, 0 tens and 5 ones make 205.
$5{,}347 \div 26 = 205 \text{ R } 17$

Answer

$5{,}347 \div 26 = 205$ R $17$ . Each of the $26$ groups gets $205$ , and $17$ are left over.

Leaving out the 0 moves the 2 from the hundreds place to the tens place

Number	Thousands	Hundreds	Tens	Ones
Dividend: $5{,}347$	5	3	4	7
Quotient: $205$	no digit	2	0	5
Wrong quotient, 0 left out: $25$	no digit	no digit	2	5

You can check a division answer by putting the dividend back together.

Dividing shared the dividend into equal groups, with some left over. To check, go the other way.

Multiply the quotient by the divisor. This is the amount in all the groups.
Add the remainder, which is the amount left over.
Compare. The total should equal the dividend. The remainder should also be less than the divisor.

When both are true, your quotient and remainder are right. If either one is not true, look for a mistake. The number sentence below shows steps 1 and 2.

(\text{quotient} \times \text{divisor}) + \text{remainder} = \text{dividend}

Checking answers from this lesson. The last row has the 0 left out, and the check catches it.

Answer to check	Multiply the quotient by the divisor	Add the remainder	Does the total equal the dividend?	Is the remainder less than the divisor?
$87 \div 21 = 4$ R $3$	$4 \times 21 = 84$	$84 + 3 = 87$	Yes	Yes, $3 < 21$
$745 \div 32 = 23$ R $9$	$23 \times 32 = 736$	$736 + 9 = 745$	Yes	Yes, $9 < 32$
$5{,}347 \div 26 = 205$ R $17$	$205 \times 26 = 5{,}330$	$5{,}330 + 17 = 5{,}347$	Yes	Yes, $17 < 26$
$5{,}347 \div 26 = 25$ R $17$ (the 0 left out)	$25 \times 26 = 650$	$650 + 17 = 667$	No, $667$ is not $5{,}347$	Yes, $17 < 26$

Worked example

A 4-digit dividend, then a check

Find $2{,}587 \div 41$ . Then check your answer.

Decide where the first digit of the quotient goes. The first two digits make 25, and 25 is less than 41. So use the first three digits: 258 tens. The first digit goes in the tens place, and the quotient will have 2 digits.
$25 < 41$
Divide the 258 tens. Each group can get 6 tens, because 6 × 41 = 246. They cannot get 7 each, because 7 × 41 = 287 is more than 258. Write 6 in the tens place and subtract.
$258 - 246 = 12$
Trade the 12 tens for 120 ones and join them with the 7 ones. Now you have 127 ones. Each group can get 3 ones, because 3 × 41 = 123. They cannot get 4 each, because 4 × 41 = 164 is more than 127. Write 3 in the ones place and subtract. The 4 left over is the remainder, so the answer is 63 R 4.
$127 - 123 = 4$
Now check. Multiply the quotient, 63, by the divisor, 41, to find the amount in all the groups. You can split it up: 60 × 41 = 2,460 and 3 × 41 = 123. Then 2,460 + 123 = 2,583.
$63 \times 41 = 2{,}583$
Add the remainder, 4. The total equals the dividend, 2,587. The remainder 4 is also less than the divisor 41. Both are true, so the answer is right.
$2{,}583 + 4 = 2{,}587$

Answer

$2{,}587 \div 41 = 63$ R $4$ . The check gives back the dividend: $63 \times 41 + 4 = 2{,}587$ , and $4$ is less than $41$ .

The big idea: Divide one place at a time. First decide where the first digit of the quotient goes. Then divide, multiply and subtract at each place, trading what is left to the next place, until you finish the ones. Every place after the first digit gets a digit of its own, and that digit is $0$ when the place cannot be divided.

To check, multiply the quotient by the divisor and add the remainder. You should get the dividend back, and the remainder should be less than the divisor.

Next, you will learn how to tell when a quotient digit you try is too large or too small, and how to find and fix division mistakes.