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3. Multiplication and division
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Stage 3

3.04 Multiplication with up to three digits

Lesson
Practice
Lesson

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We can multiply numbers by larger numbers, once we've mastered  multiplying by single digit numbers  .

Examples

Example 1

Find 6178 \times 4.

Worked Solution
Create a strategy

Use standard algorithm method for multiplication.

Apply the idea

Write the product in a vertical algorithm:

\begin{array}{c} &&6&1&7&8 \\ &\times &&&&4 \\ \hline \\ \hline \end{array}

Start from the far right. Multiplying 4 by 8, we have 32 which can be written as 3 tens and 2 units. Write 2 underneath 4 and carry the 3 to the tens column:

\begin{array}{c} &&&6&1&\text{}^3 7&8 \\ &\times &&&&&4 \\ \hline &&&&&&2\\ \hline \end{array}

Then move to the left. Multiply 4 by 7 and add the 3 to get 28+3=31. Write 1 in the tens columns and carry the 3 to the hundreds column:

\begin{array}{c} &&&6&\text{}^3 1&\text{}^3 7&8 \\ &\times &&&&&4 \\ \hline &&&&&1&2\\ \hline \end{array}

Move to the left again. Multiply 4 by 1 and add 3 to get 4+3=7. Write 7 in the hundreds column:

\begin{array}{c} &&& 6&\text{}^3 1&\text{}^3 7&8 \\ &\times &&&&&4 \\ \hline &&&&7&1&2\\ \hline \end{array}

Multiply 4 by 6 to get 24. Write 2 in the ten thousands column and 4 in the thousands column:

\begin{array}{c} &&& 6&\text{}^3 1&\text{}^3 7&8 \\ &\times &&&&&4 \\ \hline &&2&4&7&1&2\\ \hline \end{array}

6178 \times 4=24\,712

Idea summary

We can use several different methods to solve multiplication problems, but we don't always know if we will need to do any regrouping or trading. If we use an algorithm, regrouping can be done as we solve our problem.

Multiply 2 digit numbers by 2 digit numbers

Let's see how to multiply a 2 digit number by another 2 digit number.

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Examples

Example 2

Let's use the area model to find 28\times 15.

a

Find the area of each rectangle.

A rectangle with a length of 28 and a height of 15 divided into 4 rectangles. Ask your teacher for more information.
Worked Solution
Create a strategy

For each rectangle, use the formula \text{Area} = \text{length} \times \text{width}.

Apply the idea
\displaystyle \text{Area of top left rectangle}\displaystyle =\displaystyle 25\times 10Multiply the sides
\displaystyle =\displaystyle 250
\displaystyle \text{Area of top right rectangle}\displaystyle =\displaystyle 3\times 10Multiply the sides
\displaystyle =\displaystyle 30
\displaystyle \text{Area of bottom left rectangle}\displaystyle =\displaystyle 25\times5Multiply the sides
\displaystyle =\displaystyle \dfrac{1}{2}\times 25\times 10Halve 25\times 10
\displaystyle =\displaystyle \dfrac{1}{2} \times 250
\displaystyle =\displaystyle 125
\displaystyle \text{Area of bottom right rectangle}\displaystyle =\displaystyle 3\times5Multiply the sides
\displaystyle =\displaystyle 15
A rectangle with a length of 28 and a height of 15 divided into 4 rectangles. Ask your teacher for more information.
b

Find the sums of each column.

A rectangle with a length of 28 and a height of 15 divided into 4 rectangles. Ask your teacher for more information.
Worked Solution
Create a strategy

Use a vertical algorithm to add the numbers in each column.

Apply the idea

For the first column, write the numbers 250 and 125 in a vertical algorithm and add the values down each column.\begin{array}{c} &&2&5&0\\ + &&1&2&5 \\ \hline &&3&7&5 \\ \hline \end{array}

For the second column, write the numbers 30 and 15 in a vertical algorithm and add the values down each column.

\begin{array}{c} &&3&0\\ + &&1&5 \\ \hline &&4&5 \\ \hline \end{array}

A rectangle with a length of 28 and a height of 15 divided into 4 rectangles. Ask your teacher for more information.
c

What is the total area of the rectangles?

A rectangle with a length of 28 and a height of 15 divided into 2 rectangles. Ask your teacher for more information.
Worked Solution
Create a strategy

Use a vertical algorithm to add the numbers in each rectangle.

Apply the idea

Write the numbers in a vertical algorithm.

\begin{array}{c} &&3&7&5 \\ &+&&4&5 \\ \hline \\ \hline \end{array}

In the ones column 5+5=10 so we write 0 in the ones column and carry the 1 to the tens column:

\begin{array}{c} &&3&\text{}^17&5 \\ &+&&4&5 \\ \hline &&&&0 \\ \hline \end{array}

In the tens column 1+7+4=12 so we write 2 in the tens column and carry the 1 to the hundreds column:

\begin{array}{c} &&\text{}^13&\text{}^17&5 \\ &+&&4&5 \\ \hline &&&2&0 \\ \hline \end{array}

In the hundreds column 1+3=4 so we write 4 in the hundreds column:

\begin{array}{c} &&\text{}^13&\text{}^17&5 \\ &+&&4&5 \\ \hline &&4&2&0 \\ \hline \end{array}

\text{Total area}=420

d

Find 28\times15.

Worked Solution
Create a strategy

Use the total area found in part (c).

Apply the idea

The area of a rectangle with a length of 28 and a height of 15 is given by 28 \times 15. But we found this area to be 420 in part (c). So:

28\times 15=420

Idea summary

Using a standard algorithm means we can multiply larger numbers the same way, just with more steps. Once we know how to solve problems this way, there's no limit to how big the numbers could be.

Multiply large numbers by 1 and 2 digit numbers

What if we have larger numbers? Let's see how we can multiply larger numbers by 1 and 2 digit numbers.

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Examples

Example 3

Find the product of 243 \times 15.

Worked Solution
Create a strategy

Use the standard algorithm for multiplication to find the product.

Apply the idea

Set up the vertical algorithm: \begin{array}{c} &&&2&4&3 \\ &\times && &1&5 \\ \hline &&&& \\ \hline \end{array}

First we will multiply 243 by 5.

3 \times 5 = 15 so we put the 5 in the ones place and carry the 1 to the tens place.

\begin{array}{c}& &&2&{}^14&3 \\ &\times & & &1&5 \\ \hline && &&& 5 \\ \hline \end{array}

4 \times 5=20 then add the carried over 1 to get 21. Put the 1 in the tens place and carry the 2 to the hundreds place.

\begin{array}{c} &&&{}^22&{}^14&3 \\ &\times && &1&5 \\ \hline &&&&1& 5 \\ \hline \end{array}

5 \times 2=10 then add the carried over 2 to get 12. Put the 1 in the thousands place and the 2 in hundreds place.

\begin{array}{c} &&&{}^22&{}^14&3 \\ &\times && &1&5 \\ \hline &&1 &2&1&5 \\ \hline \end{array}

Now we will multiply 243 by the 1 in the tens place. We will write our answer underneath our previous answer.

Since we are multiplying by a number in the tens place we will place a 0 in the units place.

\begin{array}{c}& &&2&4&3 \\ &\times& &&1&5 \\ \hline& &1&2&1&5 \\ &&&&& 0 \\ \hline \end{array}

1\times 3=3 so we put a 3 in the tens place:

\begin{array}{c}& &&2&4&3 \\ &\times& &&1&5 \\ \hline& &1&2&1&5 \\ &&&&3& 0 \\ \hline \end{array}

1\times 4=4 so we put a 4 in the hundreds place:

\begin{array}{c}& &&2&4&3 \\ &\times& &&1&5 \\ \hline& &1&2&1&5 \\ &&&4&3& 0 \\ \hline \end{array}

1\times 2=2 so we put a 2 in the thousands place:

\begin{array}{c}& &&2&4&3 \\ &\times& &&1&5 \\ \hline& &1&2&1&5 \\ &&2&4&3& 0 \\ \hline \end{array}

Add our two answers to get the final answer:

\begin{array}{c} && &2 &4 &3 \\ &\times && &1 &5 \\ \hline &&1 &2 &1 &5 \\ &+& 2 &4 &3 & 0 \\ \hline &&3&6&4&5 \end{array}

243\times 15=3645

Idea summary

When multiplying by a two digit number using a vertical algorithm:

  • Multiply by the units digit first.

  • Put a 0 in the units column and then multiply by the tens digit.

  • Add the two products together to get the final answer.

Outcomes

MA3-6NA

selects and applies appropriate strategies for multiplication and division, and applies the order of operations to calculations involving more than one operation

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