You might notice that sometimes the standard algorithm is called the 'vertical algorithm'. Let's think about why. When we use the standard algorithm, we line our numbers up in 'vertical' place value columns.

Addition by partitioning

When we add numbers, we can break our numbers up (partition) and add the parts in a different order. Let's see how doing this helps us, in this video.

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Examples

Example 2

Let's find the value of 705 + 205, by partitioning the numbers.

Fill in the box with the missing number.

705 = 700 + ⬚

Worked Solution

Create a strategy

Put the numbers in a place value table.

Apply the idea

Hundreds	Tens	Ones
7	0	5
7	0	0

The only place where that the two numbers differ is in the units column. The first number has 5 units where as the second number has 0 units.

So we need to add 5 more to the number 700 to equal 705.

705 = 700 + 5

Fill in the box with the missing number.

205 = ⬚ + 5

Worked Solution

Create a strategy

Put the numbers in a place value table.

Apply the idea

Hundreds	Tens	Ones
2	0	5
		5

In the place value table, both 205 and 5 have 5 units but they don't match up in the tens and hundreds columns.

So we need to add 0 tens and 2 hundreds or 200 to 5 to get 205.

205 = 200 + 5

Find the value of 705 + 205.

Worked Solution

Create a strategy

Add the partitions of the two numbers.

Apply the idea

In parts (a) and (b), we have the partitions of these two numbers:

705 = 700 + 5 \\ 205 = 200 + 5

We can use these to add the two numbers.

\displaystyle 705 + 205	\displaystyle =	\displaystyle 700 + 5 + 200 + 5	Use the partitions
	\displaystyle =	\displaystyle 700 + 200 + 5 + 5	Group the hundreds and ones
	\displaystyle =	\displaystyle 900 + 10	Add the hundreds and ones
	\displaystyle =	\displaystyle 910

705 + 205 = 910

Idea summary

Addition by partitioning involves splitting numbers into each place value. The place values are then added separately.

Addition by place value columns

What happens with larger numbers? We can see in this video that writing our numbers down the page helps us with trading, or regrouping. Let's use a vertical algorithm to solve one.

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Examples

Example 3

Find the value of 6185 + 8282.

Worked Solution

Create a strategy

Use the vertical algorithm method.

Apply the idea

Write it in a vertical algorithm.\begin{array}{c} & & &6 &1 &8 &5 \\ &+ & &8 &2 &8 &2 \\ \hline & \\ \hline \end{array}

Add the units column.\begin{array}{c} & & &6 &1 &8 &5 \\ &+ & &8 &2 &8 &2 \\ \hline &&&&&6&7 \\ \hline \end{array}

Add the tens column: 8+8=16, so we bring down the 6 in the tens column and carry the 1 to the hundreds column:

\begin{array}{c} & & &6 &{}^11 &8 &5 \\ &+ & &8 &2 &8 &2 \\ \hline & & & & &6 &7 \\ \hline \end{array}

Add the hundreds column.

\begin{array}{c} & & &6 &{}^11 &8 &5 \\ &+ & &8 &2 &8 &2 \\ \hline & & & &4 &6 &7 \\ \hline \end{array}

Add the thousands column: 6 + 8 = 14. Bring down 4 to thousands column and carry on 1 to the ten thousands column.

\begin{array}{c} & &{}^1 &6 &{}^11 &8 &5 \\ &+ & &8 &2 &8 &2 \\ \hline & & &4 &4 &6 &7 \\ \hline \end{array}

So we have 1 + 0 = 1 for the ten thousands column:

\begin{array}{c} & &{}^1 &6 &{}^11 &8 &5 \\ &+ & &8 &2 &8 &2 \\ \hline & &1 &4 &4 &6 &7 \\ \hline \end{array}

So 6185 + 8282 = 14 \, 467.

Idea summary

Whenever we get more than 9 for one of the digits in our number, we have to regroup, or trade, to the next place value. All we need to remember is that we can trade 10 of any place for 1 of the places to the left.

Outcomes

ACMNA073

Apply place value to partition, rearrange and regroup numbers to at least tens of thousands to assist calculations and solve problems

ACMNA076

Develop efficient mental and written strategies and use appropriate digital technologies for multiplication and for division where there is no remainder

2.03 Addition with regrouping

Ideas

Are you ready?

Examples

Example 1

Addition by partitioning

Examples

Example 2

Addition by place value columns

Examples

Example 3

Outcomes

ACMNA073

ACMNA076

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