Add and Subtract Multi-Digit Whole Numbers | Grade 5 Math | LevelKite

When you add, you join amounts into one total. The numbers you join are addends. The total is the sum.

Picture three jars that hold $25$ , $40$ and $75$ marbles. You can pour them into one tub in any order. You can also mix any two jars first and pour in the third jar later. Each way, the tub ends up with the same $140$ marbles. An empty jar would add $0$ marbles, so it would change nothing.

These ideas are the three properties of addition. A property is a rule that is true for every number you try. These three belong to addition, and they help you make a sum easier to find. In the table, parentheses such as $(40 + 75)$ show which addends you add first.

The three properties of addition, shown with the jar numbers

Property	What it says	Number sentence	Check
Commutative property	You can change the order of the addends. The sum stays the same.	$40 + 75 = 75 + 40$	Both sides equal $115$ .
Associative property	You can change which addends you group together. The sum stays the same.	$(40 + 75) + 25 = 40 + (75 + 25)$	Left side: $115 + 25 = 140$ . Right side: $40 + 100 = 140$ .
Identity property	Adding $0$ to a number gives that same number.	$140 + 0 = 140$	You add nothing, so $140$ stays $140$ .

Worked example

Reorder and group addends

Find $175 + 0 + 68 + 25$ . Use the properties of addition to make the sum easier to find.

Find the 0 first. The identity property says that adding 0 gives the same number, so 175 + 0 is just 175. You can leave the 0 out.
$175 + 0 + 68 + 25 = 175 + 68 + 25$
Look for addends that make a number ending in zeros. 175 and 25 make 200, but 68 sits between them. The commutative property lets you change the order of addends, so write 25 before 68.
$175 + 68 + 25 = 175 + 25 + 68$
The associative property lets you choose which addends to add first. Group 175 and 25 with parentheses, and add them.
$(175 + 25) + 68 = 200 + 68$
Add the last addend. 200 has 0 tens and 0 ones, so the 6 tens and 8 ones of 68 fill those places.
$200 + 68 = 268$

Answer

The sum $175 + 0 + 68 + 25$ is $268$ .

The same properties explain how to add large numbers. Any whole number is the sum of the values of its digits. For example, $4{,}876$ is $4{,}000 + 800 + 70 + 6$ . Each part counts a different unit. In this lesson, a unit means a place value size, such as a one, a ten, a hundred or a thousand.

When you add two large numbers, the commutative and associative properties let you reorder and group all the parts. So you can put like units together: thousands with thousands, hundreds with hundreds, tens with tens, and ones with ones. That is why the written method, with one number above the other, adds one place at a time. Here is $4{,}876 + 2{,}359$ with like units grouped in parentheses:

4{,}876 + 2{,}359 = (4{,}000 + 2{,}000) + (800 + 300) + (70 + 50) + (6 + 9)

Key concept

Regrouping renames a number without changing its value

The written method adds and subtracts one place at a time, starting with the ones. Sometimes a place needs a trade. To regroup is to trade between a unit and the next greater unit. In place value, each unit is $10$ times the unit to its right, so the trade is always $10$ for $1$ .

Adding: a place can show only one digit. When a place totals $10$ or more, trade $10$ of that unit for $1$ of the next greater unit.
Subtracting: when a place has too few to take away from, trade $1$ of the next greater unit for $10$ of that unit.

The number keeps its value. Only its name changes. For example, $15$ ones is the same amount as $1$ ten and $5$ ones.

10 \text{ ones} = 1 \text{ ten} \qquad 10 \text{ tens} = 1 \text{ hundred} \qquad 10 \text{ hundreds} = 1 \text{ thousand}

Worked example

Add with regrouping in three places

Find $4{,}876 + 2{,}359$ .

Write the numbers one above the other, so each place has its own column. Start with the ones, because each regrouped unit moves one place to the left, into a column you have not added yet. 6 ones plus 9 ones is 15 ones. Regroup 15 ones as 1 ten and 5 ones. Write 5 in the ones place and move the 1 ten to the tens column.
$6 + 9 = 15$
Add the tens, and include the regrouped ten. 7 tens plus 5 tens plus 1 ten is 13 tens. Regroup 13 tens as 1 hundred and 3 tens. Write 3 in the tens place and move the 1 hundred to the hundreds column.
$7 + 5 + 1 = 13$
Add the hundreds, and include the regrouped hundred. 8 hundreds plus 3 hundreds plus 1 hundred is 12 hundreds. Regroup 12 hundreds as 1 thousand and 2 hundreds. Write 2 in the hundreds place and move the 1 thousand to the thousands column.
$8 + 3 + 1 = 12$
Add the thousands, and include the regrouped thousand. 4 thousands plus 2 thousands plus 1 thousand is 7 thousands. That is less than 10, so you do not need to regroup. Write 7 in the thousands place.
$4 + 2 + 1 = 7$
Read the digits you wrote from the greatest place to the least: 7 thousands, 2 hundreds, 3 tens and 5 ones.
$4{,}876 + 2{,}359 = 7{,}235$

Answer

The sum of $4{,}876$ and $2{,}359$ is $7{,}235$ .

The addition 4,876 + 2,359 in a place value chart

Row	Thousands	Hundreds	Tens	Ones
Regrouped unit coming in	$1$	$1$	$1$	none
$4{,}876$	$4$	$8$	$7$	$6$
$2{,}359$	$2$	$3$	$5$	$9$
Total in this place	$7$	$12$	$13$	$15$
Digit you write	$7$	$2$	$3$	$5$

You also subtract one place at a time, starting with the ones. You take ones from ones, tens from tens, and so on. The answer to a subtraction is the difference. In this lesson, the number you start with is always the greater one.

Sometimes a place has too few to take away from. Then you regroup the other way: trade $1$ of the next greater unit for $10$ of the unit you need.

Look at $452 - 138$ . In the ones place, $2$ ones are too few to take $8$ ones away. So trade $1$ ten for $10$ ones. Now $452$ has a new name: $4$ hundreds, $4$ tens and $12$ ones. It is still $452$ , because $400 + 40 + 12 = 452$ . Subtract each place: $12 - 8 = 4$ ones, $4 - 3 = 1$ ten, and $4 - 1 = 3$ hundreds. So $452 - 138 = 314$ .

Worked example

Subtract across two zeros

Find $6{,}004 - 2{,}378$ .

Start with the ones. 4 ones are too few to take 8 ones away, so you need a trade. The tens place and the hundreds place both hold 0, so they have nothing to give. Go left to the first place that is not 0, which is the thousands. Regroup 1 thousand as 10 hundreds. That leaves 5 thousands and makes 10 hundreds.
$6 \text{ thousands} + 0 \text{ hundreds} = 5 \text{ thousands} + 10 \text{ hundreds}$
Now the hundreds place has something to give. Regroup 1 hundred as 10 tens. That leaves 9 hundreds and makes 10 tens.
$10 \text{ hundreds} + 0 \text{ tens} = 9 \text{ hundreds} + 10 \text{ tens}$
Regroup 1 ten as 10 ones. That leaves 9 tens. The 10 new ones join the 4 ones to make 14 ones. The number is still 6,004, because 5,000 + 900 + 90 + 14 = 6,004. It just has a new name.
$10 \text{ tens} + 4 \text{ ones} = 9 \text{ tens} + 14 \text{ ones}$
Every place now has enough. Subtract place by place, from the ones to the thousands. 14 ones take away 8 ones leaves 6 ones. 9 tens take away 7 tens leaves 2 tens. 9 hundreds take away 3 hundreds leaves 6 hundreds. 5 thousands take away 2 thousands leaves 3 thousands.
$14 - 8 = 6 \qquad 9 - 7 = 2 \qquad 9 - 3 = 6 \qquad 5 - 2 = 3$
Read the digits from the greatest place to the least: 3 thousands, 6 hundreds, 2 tens and 6 ones.
$6{,}004 - 2{,}378 = 3{,}626$

Answer

The difference $6{,}004 - 2{,}378$ is $3{,}626$ .

The first four rows all name 6,004. Then you subtract 2,378 place by place.

Step	Thousands	Hundreds	Tens	Ones
Start with $6{,}004$	$6$	$0$	$0$	$4$
Trade 1 thousand for 10 hundreds	$5$	$10$	$0$	$4$
Trade 1 hundred for 10 tens	$5$	$9$	$10$	$4$
Trade 1 ten for 10 ones	$5$	$9$	$9$	$14$
Take away $2{,}378$	$2$	$3$	$7$	$8$
Difference	$3$	$6$	$2$	$6$

Do not flip the digits. In $6{,}004 - 2{,}378$ , a common mistake is to subtract the smaller digit from the larger digit in every column: $8 - 4 = 4$ , $7 - 0 = 7$ , $3 - 0 = 3$ and $6 - 2 = 4$ . The zeros make this tempting, because $7 - 0$ looks easy. It gives $4{,}374$ , but the difference is $3{,}626$ .

It fails because the $8$ , $7$ and $3$ belong to the number you are taking away. Flipping the ones column takes $4$ away from $8$ . The problem asks you to take $8$ ones away from $6{,}004$ . The flipped work really answers a different problem, $6{,}378 - 2{,}004$ . The commutative property is true for addition, not for subtraction. When a place has too few, regroup. Do not change the order.

Here is the idea to keep. The properties of addition let you reorder and group addends, so you can choose an easier way to add. The written methods use place value: you add or subtract one place at a time, and you regroup when a place needs it. In addition, $10$ of a unit become $1$ of the next greater unit. In subtraction, $1$ of a unit becomes $10$ of the next smaller unit, even when you must pass through zeros. Regrouping changes the name of a number, never its value.

Next, you will round numbers to estimate a sum or a difference, and use the estimate to judge whether an exact answer is reasonable. You will also find missing numbers and solve word problems.