3rd Grade · 4th Grade · 3.NBT.A.2 · 4.NBT.B.4

Subtraction with regrouping in 3rd and 4th grade

Subtraction with regrouping is what many parents learned as borrowing. When a top digit is too small to subtract from, your child trades 1 from the next place to the left for 10 in the place being worked on. Your child adds and subtracts within 1000 in 3rd grade and uses the standard written method for bigger numbers in 4th grade.

What the standards say

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

3.NBT.A.2 Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction.

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm.

What it means, in plain words

3rd grade: adding and subtracting within 1000 3.NBT.A.2

By the end of 3rd grade, your child should be quick and accurate at adding and subtracting within 1000. The work rests on place value: a digit is worth more or less depending on its place. In 452, the 4 means 4 hundreds, the 5 means 5 tens and the 2 means 2 ones.

Classes teach several ways to do three-digit addition and subtraction. Your child might work with the hundreds, tens and ones separately, count up from the smaller number or write the numbers in columns. Subtraction with regrouping in 3rd grade means trading when a column runs short: 1 ten becomes 10 ones, or 1 hundred becomes 10 tens.

Adding uses the same trade in the other direction. When a column adds up to 10 or more, 10 ones become 1 ten and move to the tens column. Your child can check a subtraction by adding: the answer plus the number taken away gives the starting number.

Your child should be able to:

Add and subtract three-digit numbers quickly and correctly
Trade 1 ten for 10 ones, or 1 hundred for 10 tens
Use more than one method, such as counting up or working in columns
Check a subtraction by adding

Example

What is 452 minus 176?

Ones: 2 is less than 6, so trade 1 ten for 10 ones. Now there are 12 ones, and $12 - 6 = 6$ .
Tens: 4 tens are left, and 4 is less than 7. Trade 1 hundred for 10 tens. Now there are 14 tens, and $14 - 7 = 7$ .
Hundreds: 3 hundreds are left, and $3 - 1 = 2$ .
So 452 minus 176 is 276. Check: $276 + 176 = 452$ .

Example

What is 358 plus 267?

Ones: $8 + 7 = 15$ . Write 5 and move 1 ten to the tens column.
Tens: $5 + 6 + 1 = 12$ tens. Write 2 and move 1 hundred to the hundreds column.
Hundreds: $3 + 2 + 1 = 6$ .
So 358 plus 267 is 625.

4th grade: the standard algorithm for bigger numbers 4.NBT.B.4

Subtraction with regrouping in 4th grade uses the standard algorithm, the usual written method: write the numbers in columns, lined up by place, and work from right to left. The numbers grow into the thousands and beyond, up to 1,000,000. This is the one fluency 4th grade requires, so by the end of the year it should feel quick and sure.

The standard algorithm for addition works the same way. Your child adds each column from right to left and carries 1 to the next column when a column reaches 10 or more. Multi-digit addition with three numbers at once uses the same steps.

The hardest case is a zero on top. In 5,002 minus 1,746, the ones need a trade, but the tens and the hundreds have nothing to give. Your child trades from the thousands instead: 5,002 is 500 tens and 2 ones, and trading 1 ten leaves 499 tens and 12 ones.

Your child should be able to:

Add multi-digit numbers with the standard algorithm, carrying when needed
Subtract multi-digit numbers, regrouping across several places
Regroup across zeros, as in 5,002 minus 1,746
Line up the digits by place before starting

Example

What is 4,786 plus 2,569?

Ones: $6 + 9 = 15$ . Write 5 and carry 1.
Tens: $8 + 6 + 1 = 15$ . Write 5 and carry 1.
Hundreds: $7 + 5 + 1 = 13$ . Write 3 and carry 1.
Thousands: $4 + 2 + 1 = 7$ .
So 4,786 plus 2,569 is 7,355.

Example

What is 5,002 minus 1,746?

The ones need a trade, but the tens and hundreds are 0.
Trade from the thousands: 5,002 becomes 4 thousands, 9 hundreds, 9 tens and 12 ones.
Ones: $12 - 6 = 6$ . Tens: $9 - 4 = 5$ . Hundreds: $9 - 7 = 2$ . Thousands: $4 - 1 = 3$ .
So 5,002 minus 1,746 is 3,256. Check: $3{,}256 + 1{,}746 = 5{,}002$ .

Where children usually get stuck

For 523 minus 258, your child writes 335 instead of 265.
In each column the child takes the smaller digit from the bigger one, whichever is on top. That skips the trade, and each column on its own looks fine.
What helps. Read the problem as a story: you have 523 and give away 258. Ask whether you can give away 8 ones when you have only 3. Pennies and dimes make the trade easy to see.
For 403 minus 156, your child changes the 4 to a 3, writes 13 ones and leaves the 0 tens alone. The answer comes out as 257 instead of 247.
The ten was taken straight from the hundreds, skipping the tens place. When the tens place is 0, 1 hundred has to become 10 tens first, and then 1 of those tens becomes 10 ones.
What helps. Say the number in tens together: 403 is 40 tens and 3 ones. Trading 1 ten leaves 39 tens and 13 ones, which is why a 3 and a 9 go over the 4 and the 0.
To add 1,258 and 47, your child writes the 47 under the 2 and the 5 and gets 1,728.
The numbers were lined up on the left, the way words line up on a page. Digits have to line up by place, ones under ones.
What helps. Use graph paper with one digit in each box. Write both ones digits in the same column first. The sum is 1,305.

Try four practice questions

Real 3rd 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
Find the sum.
$\displaystyle \begin{array}{r} 246 \\ +\,37 \\ \hline \end{array}$
A
283
B
293
C
273
D
284
Show the answer and how to solve it
The sum is $283$ . $13$ ones are $1$ ten and $3$ ones. Regroup that $1$ ten into the tens column.
1. 1Look at the addition. The numbers $246$ and $37$ are stacked by place value.
2. 2Add the ones digits. $6 + 7 = 13$ . $10$ ones make $1$ ten. $13$ ones are $1$ ten and $3$ ones. Write $3$ in the ones place and regroup $1$ ten into the tens column. Then add the tens digits and the regrouped $1$ : $4 + 3 + 1 = 8$ . Write $8$ in the tens place. Finally, add the hundreds digits: $2 + 0 = 2$ . Write $2$ in the hundreds place.
3. 3The completed addition shows the sum.
Common mistake. Do not leave out the regrouped ten. $13$ ones are $1$ ten and $3$ ones. Include that $1$ ten in the tens total: $4 + 3 + 1 = 8$ .
The idea underneath. $10$ ones make $1$ ten. $13$ ones are $1$ ten and $3$ ones. Regroup that $1$ ten into the tens column.
2On level
A student added two whole numbers using the standard algorithm. The student's completed work is shown below.
Select the statement that best describes the student's work.
A
The student added the carried 1 to the tens column twice.
B
The student carried an extra 1 from the tens column to the hundreds column.
C
The student carried the 1 to the hundreds column instead of the tens column.
D
The student forgot to add the carried 1 to the tens column.
Show the answer and how to solve it
The correct sum of $3{,}417 + 2{,}568$ is 5,985, but the displayed work shows 5,995 — exactly 10 too high. The tens digit is 9 instead of 8, which means the carried 1 was added to the tens column twice instead of once.
1. 1Look at the student's completed work. The student added $3{,}417 + 2{,}568$ and wrote 5,995 as the sum. We need to check whether this is correct, and if not, identify the specific error.
2. 2Compute the correct sum. Ones: $7 + 8 = 15$ , write 5 and carry 1. Tens: $1 + 6 + 1 (carried) = 8$ . Hundreds: $4 + 5 = 9$ . Thousands: $3 + 2 = 5$ . The correct sum is 5,985.
3. 3Compare the correct sum to the student's sum. The correct sum is 5,985, but the work shows 5,995. The student's answer is 10 more than correct, so the error must be in the tens column.
4. 4The correct tens digit is $1 + 6 + 1 (carried) = 8$ . The student wrote 9 — one more than expected. This means the carried 1 was added to the tens column an extra time, so it was counted twice instead of once.
5. 5The statement that best describes this work is: 'The student added the carried 1 to the tens column twice.'
Common mistake. A common mistake is assuming any error must be a forgotten carry. Here the sum is 10 too high, not too low, so the carry was used one extra time. Forgetting the carry would produce 5,975 with a 7 in the tens place, not the 9 shown.
The idea underneath. When checking another student's addition, compare each column's digit one at a time. The position of the wrong digit tells you which column has the error, and whether the sum is too high or too low tells you whether a carry was duplicated, forgotten, or misplaced.
3On level
$\displaystyle \begin{array}{r} 143 \\ +\,128 \\ \hline 261 \end{array}$
A student wrote this work:
Ones: $3 + 8 = 11$ . Write $1$ and regroup $1$ to the next column.
Tens: $4 + 2 = 6$ . Write $6$ .
Hundreds: $1 + 1 = 2$ .
Which statement best describes the work?
A
The student used 2 instead of 1 when regrouping.
B
The student left out the regrouped 1.
C
The student added all the numbers correctly.
D
The student added the ones digits incorrectly.
Show the answer and how to solve it
The tens calculation is $4 + 2 = 6$ . It leaves out the $1$ regrouped from the previous column. The student left out the regrouped 1.
1. 1Add the ones: $3 + 8 = 11$ . Write $1$ and regroup $1$ to the next column.
2. 2Add the tens: $4 + 2 + 1 = 7$ . Write $7$ .
3. 3Add the hundreds: $1 + 1 = 2$ . The correct sum is $271$ .
4. 4The tens calculation is $4 + 2 = 6$ . It leaves out the $1$ regrouped from the previous column.
Common mistake. Check the written calculations before naming a mistake. A difference of 10 or 100 can also come from adding a column incorrectly.
The idea underneath. Check each column and how the regrouped 1 is used. A wrong total alone does not show how the mistake happened.
4Stretch
Complete the subtraction. Fill in the missing digits in the boxes.
Regrouped:
13
12
7 5 3 6 2
− 2 8 5 4 7
4 6
1 5
Show the answer and how to solve it
The difference is $75{,}362 - 28{,}547 = 46{,}815$ . The missing values are: regrouped ten-thousands = $6$ , regrouped thousands = $14$ , regrouped tens = $5$ , and hundreds difference = $8$ .
1. 1Start in the ones column. The top digit is 2 and the bottom digit is 7. Since 2 is less than 7, regroup before subtracting.
2. 2Borrow 1 ten from the tens column. The tens digit changes from 6 to 5, and the ones digit becomes 12. Subtract in the ones column.
3. 3Move to the tens column. The new top digit is 5 and the bottom digit is 4. No regrouping needed.
4. 4In the hundreds column, the top digit is 3 and the bottom digit is 5. Since 3 is less than 5, regroup. Borrow 1 thousand from the thousands column. The thousands digit changes from 5 to 4, and the hundreds digit becomes 13.
5. 5In the thousands column, the new top digit is 4 and the bottom digit is 8. Since 4 is less than 8, regroup again. Borrow 1 ten-thousand from the ten-thousands column. The ten-thousands digit changes from 7 to 6, and the thousands digit becomes 14.
6. 6In the ten-thousands column, the new top digit is 6 and the bottom digit is 2. Subtract.
7. 7The completed subtraction reads 75,362 − 28,547 = 46,815. The missing values are: regrouped ten-thousands = 6, regrouped thousands = 14, regrouped tens = 5, and hundreds difference = 8.
Common mistake. A common mistake is writing the regrouped thousands as $15$ by adding $10$ to the original $5$ . The thousands column must first lend $1$ to the hundreds (reducing from $5$ to $4$ ), and only then borrow $10$ from the ten-thousands to become $14$ . Both steps must be tracked in order.
The idea underneath. Each regrouping is really two moves at once: take 1 from the column to your left (reducing it by 1) and add 10 to the current column. Doing only one of these moves changes the size of the number, so both must happen together.

Regrouped:			13		12
	7	5	3	6	2
−	2	8	5	4	7
	4	6		1	5