Topics
4. Multiplication and division
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AustraliaNSW
Stage 3

4.01 Multiplication

Lesson
Practice
Lesson

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When we need to solve multiplication problems, we have many methods we could use. Which method we choose depends on our problem, how many digits our numbers have, and the method we are more comfortable with.

Idea summary

Methods used in multiplication:

  • repeated addition

  • double-double

  • area model

  • partitioning by place value

  • partitioning by factors

  • vertical multiplication

Multiplication review

Let's look at a video on how to multiply two 2 digit numbers.

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Examples

Example 1

Calculate 26 \times 39 by completing an area model.

a

Firstly, complete the missing values in the area model.

\times206
30
9
Worked Solution
Create a strategy

Multiply the number at the top of the column by the number on the left of the row.

Apply the idea
\displaystyle 30\times 20\displaystyle =\displaystyle 600Find 3\times 2 then add two zeros
\displaystyle 30\times 6\displaystyle =\displaystyle 180Find 3\times 6 then add a zero
\displaystyle 9\times 20\displaystyle =\displaystyle 180Find 9\times 2 then add a zero
\displaystyle 9\times 6\displaystyle =\displaystyle 54
\times206
30600180
918054
b

Now add the values you found in part (a) to calculate 26\times 39.

Worked Solution
Create a strategy

Use a vertical algorithm to add the results.

Apply the idea

Write the results in a vertical algorithm and add the digits from the units column to the hundreds column:\begin{array}{c} && &{}^2 6 &0 &0 \\ && &1 &8 &0 \\ && &1 &8 &0 \\ &+& & &5 &4 \\ \hline &&1 &0 &1 &4 \\ \hline \end{array}

\displaystyle 26\times 39\displaystyle =\displaystyle 1014
Idea summary

We can use area models to multiply 2 two digit numbers.

Multiply 3 digit numbers

This video looks at using the algorithm to solve a multiplication problem.

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Examples

Example 2

Find 7\times 583.

Worked Solution
Create a strategy

Use the standard algorithm method for multiplication.

Apply the idea

Write the product in a vertical algorithm:

\begin{array}{c} &&&5&8&3 \\ &\times &&&&7 \\ \hline \\ \hline \end{array}

Start from the far right. Multiply to get 7\times 3=21. So we write the 1 in the units place and carry the 2 to the tens place:

\begin{array}{c} &&&5&{}^28&3 \\ &\times &&&&7 \\ \hline &&&&&1 \\ \hline \end{array}

Move left and multiply to get 7\times 8=56. Then we add the 2 to get 58. So we write the 8 in the tens place and carry the 5 to the hundreds place:

\begin{array}{c} &&&{}^55&{}^28&3 \\ &\times &&&&7 \\ \hline &&&&8&1 \\ \hline \end{array}

Move left and multiply to get 7\times 5=35. Then we add the 5 to get 40. So we write the 4 in the thousands place and the 0 in the hundreds place:

\begin{array}{c} &&&{}^55&{}^28&3 \\ &\times &&&&7 \\ \hline &&4&0&8&1 \\ \hline \end{array}

583\times7=4081

Idea summary

There isn't one perfect way to solve multiplication problems. The best approach is to have a range of methods to choose from, so you can choose the method that suits the problem. This is just like having a toolkit with different tools.

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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