Multiplication: Whole Numbers

Multiplication can be thought of as repeated addition. So, if you multiply a number $a$ by another number $b$ , this is the same as adding the number $a$ over and over again $b$ times. (Or adding $b$ over and over again $a$ times). For example:

$\begin{array}{l}3\times 5=5+5+5=15\\ 3\times 5=3+3+3+3+3=15\end{array}$

Another way to think of whole number multiplication $a\times b$ is to visualize objects arranged in a rectangle, with $a$ rows and $b$ columns.

$3\times 5$

Note that there are $15$ dots in the figure.

The Standard Algorithm

To multiply a multi-digit number by a one-digit number using the standard algorithm, write the two numbers on top of each other, with the ones digits vertically aligned and the multi-digit number on top.

$\begin{array}{l}127\\ \underset{\_}{\times 3}\end{array}$

Multiply the ones digit of the top number by the bottom number. Write down the ones digit of the result. If the result is greater than $10$ , carry the tens digit, as you do when adding.

Here, $7\times 3=21$ , so

$\begin{array}{l}1\stackrel{2}{2}7\\ \underset{\_}{\times 3}\\ 1\end{array}$

Now multiply the tens digit of the top number by the bottom number, and add the carried digit to the result. Here, $2\times 3=6$ , and then we add $2$ to get $8$ . Since $8$ is less than $10$ , we don't have to carry this time.

$\begin{array}{l}1\stackrel{2}{2}7\\ \underset{\_}{\times 3}\\ 81\end{array}$

Finally, multiply the hundreds digit of the top number by the bottom number. Here, $3\times 1=3$ .

$\begin{array}{l}1\stackrel{2}{2}7\\ \underset{\_}{\times 3}\\ 381\end{array}$

So, $127\times 3=381$ .

To multiply two multi-digit numbers , write the number with more digits on top. For example, to multiply $29$ by $543$ , we write

$\begin{array}{l}543\\ \underset{\_}{\times 29}\end{array}$

First multiply the top number by the ones digit of the bottom number, as explained above. $3\times 9=27$ , so write down the $7$ and carry the $2$ :

$\begin{array}{l}5\stackrel{2}{4}3\\ \underset{\_}{\times 29}\\ 7\end{array}$

$4\times 9$ is 36, plus $2$ is $38$ , so write down the 8 and carry the $3$ :

$\begin{array}{l}\stackrel{3}{5}\stackrel{2}{4}3\\ \underset{\_}{\times 29}\\ 87\end{array}$

$5\times 9$ is $45$ , plus $3$ is $48$ . There are no more digits to carry, so write down $48$ .

$\begin{array}{l}\stackrel{2}{5}\stackrel{2}{4}3\\ \underset{\_}{\times 29}\\ 4887\end{array}$

Next, we need to multiply the top number by the tens digit of the bottom number. Since we're actually multiplying by $20$ , not by $2$ , we write down a $0$ as a place holder.

$\begin{array}{l}543\\ \underset{\_}{\times 29}\\ 4887\\ 0\end{array}$

$3\times 2$ is $6$ , so write down a $6$ .

$\begin{array}{l}543\\ \underset{\_}{\times 29}\\ 4887\\ 60\end{array}$

$4\times 2$ is $8$ , so write down an $8$ .

$\begin{array}{l}543\\ \underset{\_}{\times 29}\\ 4887\\ 860\end{array}$

$5\times 2$ is $10$ , and there are no more digits to carry, so write down the $10$ .

$\begin{array}{l}543\\ \underset{\_}{\times 29}\\ 4887\\ 10860\end{array}$

The final step is to add the two results.

$\begin{array}{l}543\\ \underset{\_}{\times 29}\\ 4887\\ \underset{\_}{+10860}\\ 13947\end{array}$

So, $543\times 29=13947$ .

Like addition, multiplication is commutative for real numbers (that is, $a\times b=b\times a$ ; order does not matter) and associative (that is, $(a\times b)\times c=a\times (b\times c)$ ; grouping does not matter.) See Properties of Multiplication for more.