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Properties of Multiplication

When we multiply by a number with two or more digits, we multiply by each of the digits separately, working from right to left. Each separate product of the digits is called a partial product. When we write partial products, we must make sure to line up the ...

Multiply Whole Numbers

In order to multiply without using models, you need to know all the one digit multiplication facts. Make sure you know them fluently before proceeding in this tutorial.

multiplication-square

Image credit: Amazon UK

The table below shows the multiplication facts. Each box shows the product of the number down the left column and the number across the top row. If you are unsure about a product, model it. It is important that you memorize any number facts you do not already know so you will be ready to multiply larger numbers.

× 0 1 2 3 4 5 6 7 8 9
0 0 0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7 8 9
2 0 2 4 6 8 10 12 14 16 18
3 0 3 6 9 12 15 18 21 24 27
4 0 4 8 12 16 20 24 28 32 36
5 0 5 10 15 20 25 30 35 40 45
6 0 6 12 18 24 30 36 42 48 54
7 0 7 14 21 28 35 42 49 56 63
8 0 8 16 24 32 40 48 56 64 72
9 0 9 18 27 36 45 54 63 72 81

What happens when you multiply a number by zero? You can see that the product of any number and zero is zero. This is called the Multiplication Property of Zero.

Multiplication Property of Zero

The product of any number and \(0\) is \(0.\)

\(\begin{array}{}a·0=0\hfill \\ 0·a=0\end{array}\)

Example

Multiply:

  1. \(0·11\)
  2. \(\left(42\right)0\)

Solution

1. \(0·11\)
The product of any number and zero is zero. \(0\)
2. \(\left(42\right)0\)
Multiplying by zero results in zero. \(0\)

Extra:

Find each product:

  1. \(0·19\)
  2. \(\left(39\right)0\)

Solution

  1. \(0\)
  2. \(0\)

Extra:

Find each product:

  1. \(0·24\)
  2. \(\left(57\right)0\)

Solution

  1. \(0\)
  2. \(0\)

What happens when you multiply a number by one? Multiplying a number by one does not change its value. We call this fact the Identity Property of Multiplication, and \(1\) is called the multiplicative identity.

Identity Property of Multiplication

The product of any number and \(1\) is the number.

\(\begin{array}{c}1·a=a\\ a·1=a\end{array}\)

Example

Multiply:

  1. \(\left(11\right)1\)
  2. \(1·42\)

Solution

1. \(\left(11\right)1\)
The product of any number and one is the number. \(11\)
2. \(1·42\)
Multiplying by one does not change the value. \(42\)

Extra:

Find each product:

  1. \(\left(19\right)1\)
  2. \(1·39\)

Solution

  1. \(19\)
  2. \(39\)

Extra:

Find each product:

  1. \(\left(24\right)\left(1\right)\)
  2. \(1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}57\)

Solution

  1. \(24\)
  2. \(57\)

Earlier in this tutorial, we learned that the Commutative Property of Addition states that changing the order of addition does not change the sum. We saw that \(8+9=17\) is the same as \(9+8=17.\)

Is this also true for multiplication? Let’s look at a few pairs of factors.

\(4·7=28\phantom{\rule{2em}{0ex}}7·4=28\)

\(9·7=63\phantom{\rule{2em}{0ex}}7·9=63\)

\(8·9=72\phantom{\rule{2em}{0ex}}9·8=72\)

When the order of the factors is reversed, the product does not change. This is called the Commutative Property of Multiplication

Commutative Property of Multiplication:

Changing the order of the factors does not change their product.

\(a·b=b·a\)

Example

Multiply:

  1. \(8·7\)
  2. \(7·8\)

Solution

1. \(8·7\)
Multiply. \(56\)
2. \(7·8\)
Multiply. \(56\)

Changing the order of the factors does not change the product.

Extra:

Multiply:

  1. \(9·6\)
  2. \(6·9\)

Solution

54 and 54; both are the same.

Extra:

Multiply:

  1. \(8·6\)
  2. \(6·8\)

Solution

48 and 48; both are the same.

To multiply numbers with more than one digit, it is usually easier to write the numbers vertically in columns just as we did for addition and subtraction.

\(\begin{array}{c}\hfill 27\phantom{\rule{0.1em}{0ex}}\\ \hfill \underset{\text{___}}{×3}\end{array}\)

We start by multiplying \(3\) by \(7.\)

\(3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}7=21\)

We write the \(1\) in the ones place of the product. We carry the \(2\) tens by writing \(2\) above the tens place.

No Alt Text

Then we multiply the \(3\) by the \(2,\) and add the \(2\) above the tens place to the product. So \(3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2=6,\) and \(6+2=8.\) Write the \(8\) in the tens place of the product.

No Alt Text

The product is \(81.\)

When we multiply two numbers with a different number of digits, it’s usually easier to write the smaller number on the bottom. You could write it the other way, too, but this way is easier to work with.

Example

Multiply: \(15·4.\)

Solution

Write the numbers so the digits \(5\) and \(4\) line up vertically. \(\begin{array}{r}
15 \\
\underline{×\;4} \\
\end{array}\)
Multiply \(4\) by the digit in the ones place of \(15.\) \(4\cdot 5=20.\)
Write \(0\) in the ones place of the product and carry the \(2\) tens. \(\begin{array}{r}
\overset{2}{1}5 \\
×\;4 \\
\hline
0 \\
\end{array}\)
Multiply \(4\) by the digit in the tens place of \(15.\) \(4\cdot 1=4\).
Add the \(2\) tens we carried. \(4+2=6\).
Write the \(6\) in the tens place of the product. \(\begin{array}{r}
\overset{2}{1}5 \\
×\;4 \\
\hline
60 \\
\end{array}\)

Extra:

Multiply: \(64·8.\)

Solution

\(512\)

Extra:

Multiply: \(57·6.\)

Solution

\(342\)

Example

Multiply: \(286·5.\)

Solution

Write the numbers so the digits \(5\) and \(6\) line up vertically. \(\begin{array}{r}
286 \\
\underline{×\;5} \\
\end{array}\)
Multiply \(5\) by the digit in the ones place of \(286.\) \(5\cdot 6=30.\)
Write the \(0\) in the ones place of the product and carry the \(3\) to the tens place. Multiply \(5\) by the digit in the tens place of \(286.\) \(5\cdot 8=40\). \(\begin{array}{r}
2\overset{3}{8}6 \\
×\;5 \\
\hline
0 \\
\end{array}\)
Add the \(3\) tens we carried to get \(40+3=43\).
Write the \(3\) in the tens place of the product and carry the 4 to the hundreds place.		 \(\begin{array}{r}
\overset{4}{2}\overset{3}{8}6 \\
×\;5 \\
\hline
30 \\
\end{array}\)
Multiply \(5\) by the digit in the hundreds place of \(286.\) \(5\cdot 2=10.\)
Add the \(4\) hundreds we carried to get \(10+4=14.\)
Write the \(4\) in the hundreds place of the product and the \(1\) to the thousands place.		 \(\begin{array}{r}
\overset{4}{2}\overset{3}{8}6 \\
×\;5 \\
\hline
1,430 \\
\end{array}\)

Extra:

Multiply: \(347·5.\)

Solution

\(1,735\)

Extra:

Multiply: \(462·7.\)

Solution

\(3,234\)

When we multiply by a number with two or more digits, we multiply by each of the digits separately, working from right to left. Each separate product of the digits is called a partial product. When we write partial products, we must make sure to line up the place values.

Optional Video: Stick Multiplication and Partial Products

This video below by Mathispower4u explains how to perform stick multiplication and shows how it relates to the method of partial products.

This lesson is part of:

Introducing Numbers

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