3rd Grade · 3.OA.B.5

The commutative property of multiplication in 3rd grade

The commutative property of multiplication says you can multiply two numbers in either order and get the same answer. For example, $4 \times 6$ and $6 \times 4$ are both 24. In 3rd grade, your child uses this rule and two other properties of multiplication as shortcuts for facts that are not memorized yet.

What the standard says

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

3.OA.B.5 Apply properties of operations as strategies to multiply and divide.

What it means, in plain words

A property is a rule that works for every number. The commutative property is the turn-around rule. An array shows why it works. An array is a set of objects in equal rows, like muffins in a pan. Four rows of 6 muffins make 24. Turn the pan on its side and you see 6 rows of 4. There are still 24 muffins.

The commutative property of multiplication is useful right away. Once your child knows $3 \times 9 = 27$ , the fact $9 \times 3 = 27$ needs no extra learning. That cuts the number of facts to learn almost in half.

Your child meets two more properties this year, often without their formal names. The associative property says that with three numbers to multiply, you can start with any two of them. The distributive property breaks a hard fact into two easy ones. For example, $4 \times 9$ is $4 \times 5$ plus $4 \times 4$ , which is $20 + 16 = 36$ .

Your child should be able to:

Explain with an array why $4 \times 6$ and $6 \times 4$ have the same answer
Use a known fact, such as $3 \times 9 = 27$ , to find its turn-around fact
Choose an easy pair to multiply first when there are three numbers
Break a hard fact into two easier facts and add the answers

Example

A sticker sheet has 3 rows of 8 stickers. Another sheet has 8 rows of 3 stickers. Which sheet has more stickers?

The first sheet has $3 \times 8 = 24$ stickers.
Turn the second sheet on its side. Its 8 rows of 3 become 3 rows of 8.
So the second sheet also has $8 \times 3 = 24$ stickers.
Both sheets have 24 stickers, so neither sheet has more.

Example

What is $5 \times 8 \times 2$ ? Find it with the easiest numbers.

Change the order so the 5 and the 2 sit together: $5 \times 2 \times 8$ .
Multiply that pair first: $5 \times 2 = 10$ .
Then $10 \times 8 = 80$ .
The answer is 80. Going left to right, $40 \times 2$ , gives 80 too, with harder numbers.

Where children usually get stuck

Your child knows $3 \times 9 = 27$ but counts up by 3s, nine times, to find $9 \times 3$ .
The facts were learned as a list of separate sentences. Nobody showed that the two facts describe the same array, turned on its side.
What helps. Lay out 3 rows of 9 dried beans on a plate. Turn the plate a quarter turn and ask how many rows there are now and how many beans.
Your child writes $5 \div 15 = 3$ because $15 \div 5 = 3$ .
The turn-around rule worked for multiplication, so it gets used for division too. Division has no turn-around rule.
What helps. Act out both with crackers. 15 crackers shared by 5 people is 3 each. 5 crackers shared by 15 people is less than one cracker each.
To find $4 \times 9$ , your child writes $4 \times 5 + 4$ and gets 24.
Splitting 9 into 5 and 4 was right, but only the 5 got multiplied. The 4 stayed a plain 4 instead of 4 groups of 4.
What helps. Draw 4 rows of 9 dots with a line after the fifth dot in each row. Count each side as its own small array: 4 rows of 5 and 4 rows of 4.

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
Jaden used the break-apart strategy to solve a division problem.
$\displaystyle \begin{array}{l} 42 \div 3 = (30 \div 3) + (12 \div 3) \\ = 15 + 4 \\ = 19 \end{array}$
Select the statement that best describes Jaden's work.
A
The second division answer is computed incorrectly.
B
All steps in the computation are correct.
C
The first division answer is computed incorrectly.
D
The final sum is computed incorrectly.
Show the answer and how to solve it
Jaden's first division answer is wrong: he wrote $15$ for $30 \div 3$ , but $30 \div 3 = 10$ . The second division answer ( $4$ ) and the addition $15 + 4 = 19$ are arithmetically fine, so the error is specifically in the first division answer.
1. 1Check that the original split is valid. The first line shows $42 \div 3 = (30 \div 3) + (12 \div 3)$ . Verify the dividend: $30 + 12 = 42$ . The split is valid.
2. 2Check the first division answer. The work shows $30 \div 3 = 15$ . Confirm using multiplication: $3 \times 15 = 45$ , not $30$ . The correct value is $30 \div 3 = 10$ . The displayed $15$ is wrong.
3. 3Check the second division answer. The work shows $12 \div 3 = 4$ . Confirm using multiplication: $3 \times 4 = 12$ . The second division answer is correct.
4. 4Check the final addition. The work shows $15 + 4 = 19$ . The addition is arithmetically correct given those inputs. But one of those inputs — the $15$ — came from an incorrect division answer. The error is upstream, in the first division answer, not in the final addition itself.
Common mistake. A common mistake is to focus on whether the final addition is correct, missing that one of the division answers is computed incorrectly. Here $15 + 4 = 19$ is correct addition, but the first division answer $30 \div 3$ should be $10$ , not $15$ .
The idea underneath. In a multi-step computation, an error can hide one step earlier than it first appears. Always verify each line independently rather than trusting the final value.
2On level
Marcus is finding $12 \div 1$ . Marcus's answer is shown below.
$\displaystyle 12 \div 1 = 1$
Which calculation could explain Marcus's answer?
A
Subtracting 1 from 12.
B
Adding 1 to 12.
C
Dividing 12 by 1 correctly.
D
Dividing 12 by itself instead of by 1.
Show the answer and how to solve it
Dividing 12 by itself instead of by 1. This gives 1, which matches Marcus's answer. The correct quotient is 12.
1. 1Marcus's written answer is 1. Check it against 12 ÷ 1.
2. 2Put 12 objects in 1 group. All 12 objects are in that group, so 12 ÷ 1 = 12.
3. 3The calculation 12 ÷ 12 = 1 gives the shown answer, 1.
4. 4So dividing 12 by itself instead of by 1 could explain Marcus's answer. It gives 1, while dividing by 1 gives 12.
Common mistake. Dividing by 1 and dividing by itself are different: 12 ÷ 1 = 12, but 12 ÷ 12 = 1.
The idea underneath. Dividing into 1 group puts the whole amount in that group. Dividing by 1 keeps the number the same.
3On level
Find the missing product.
$\displaystyle 7 \times 8 = 7 \times 5 + 7 \times 3 = \underline{\phantom{000}}$
A
35
B
56
C
64
D
63
Show the answer and how to solve it
The product is $56$ , found by adding the partial products: $7 \times 5 = 35$ and $7 \times 3 = 21$ , for a total of $35 + 21 = 56$ .
1. 1The equation breaks $7 \times 8$ into two smaller products: $7 \times 5$ and $7 \times 3$ . Find each partial product, then add them.
2. 2First, compute the first partial product.
3. 3Next, compute the second partial product.
4. 4Add the two partial products to get the full answer.
Common mistake. A common mistake is reporting only one partial product. Computing $7 \times 5 = 35$ and stopping there gives $35$ — but the full product needs both partials added: $35 + 21 = 56$ .
The idea underneath. Breaking apart works because $7 \times 8$ is the same as $7 \times 5 + 7 \times 3$ when $5 + 3 = 8$ . Each partial product is easier to recall than the full fact, but the partials always add up to the same total.
4Stretch
Devontae wants to find $7 \times 8$ using a break-apart strategy. His work is shown below.
$\displaystyle 7 \times 8 = 7 \times (3 + 6) = 7 \times 3 + 7 \times 6$
Select the statement that best describes Devontae's work.
A
The student split the second factor incorrectly.
B
The student wrote subtraction instead of addition.
C
The student used the wrong factor in one partial product.
D
The reasoning is fully correct.
Show the answer and how to solve it
The student split the second factor incorrectly. The parts add to $9$ , not $8$ . The later multiplication uses those wrong parts correctly, but the split already changed the total.
1. 1Look at the break-apart work.
2. 2 $3 + 6 = 9$ . This is not $8$ , so the second factor was split incorrectly.
3. 3Splitting each of $7$ groups of $9$ into $3$ and $6$ gives $7$ groups of $3$ plus $7$ groups of $6$ . But the starting groups need $8$ each, so this split changes the total.
4. 4The parts add to $9$ , not $8$ . The later multiplication uses those wrong parts correctly, but the split already changed the total.
5. 5The best description is "The student split the second factor incorrectly."
Common mistake. Check the split before multiplying: $3 + 6 = 9$ , not $8$ .
The idea underneath. Splitting each of $7$ groups of $9$ into $3$ and $6$ gives $7$ groups of $3$ plus $7$ groups of $6$ . But the starting groups need $8$ each, so this split changes the total.