We also looked at how to use a vertical algorithm with trading, or regrouping. For addition, we regroup to the higher place value, but for subtraction, we regroup down to the next lower place value.

Examples

Example 1

Find the value of 34\,246+3213.

Worked Solution

Create a strategy

Use the standard algorithm method.

Apply the idea

Write it in a vertical algorithm.\begin{array}{c} & &3 &4 &2 &4 &6 \\ &+ & &3 &2 &1 &3 \\ \hline & \\ \hline \end{array}

Add the units column first: 6 + 3 = 9.

\begin{array}{c} & &3 &4 &2 &4 &6 \\ &+ & &3 &2 &1 &3 \\ \hline & & & & & &9 \\ \hline \end{array}

Add the tens column: 4 + 1 = 5

\begin{array}{c} & &3 &4 &2 &4 &6 \\ &+ & &3 &2 &1 &3 \\ \hline & & & & &5 &9 \\ \hline \end{array}

Add the hundreds column: 2 + 2 = 4

\begin{array}{c} & &3 &4 &2 &4 &6 \\ &+ & &3 &2 &1 &3 \\ \hline & & & &4 &5 &9 \\ \hline \end{array}

Add the thousands column: 4 + 3 = 7

\begin{array}{c} & &3 &4 &2 &4 &6 \\ &+ & &3 &2 &1 &3 \\ \hline & & &7 &4 &5 &9 \\ \hline \end{array}

Add the ten thousands column: 3 + 0 = 3

\begin{array}{c} & &3 &4 &2 &4 &6 \\ &+ & &3 &2 &1 &3 \\ \hline & & 3 &7 &4 &5 &9 \\ \hline \end{array}

34\,246 + 3213 = 37\,459

Example 2

Find the value of 49\,481 + 49\,100.

Worked Solution

Create a strategy

Use the standard algorithm method.

Apply the idea

Write it in a vertical algorithm.\begin{array}{c} & & &4 &9 &4 &8 &1 \\ &+ & &4 &9 &1 &0 &0 \\ \hline & \\ \hline \end{array}

Add the units column first.\begin{array}{c} & & &4 &9 &4 &8 &1 \\ &+ & &4 &9 &1 &0 &0 \\ \hline & & & & & & &1\\ \hline \end{array}

In the tens column, we get 8+0=8.\begin{array}{c} & & &4 &9 &4 &8 &1 \\ &+ & &4 &9 &1 &0 &0 \\ \hline & & & & & &8 &1\\ \hline \end{array}

In the hundreds column, we get 4+1=5.\begin{array}{c} & & &4 &9 &4 &8 &1 \\ &+ & &4 &9 &1 &0 &0 \\ \hline & & & & & 5&8 &1\\ \hline \end{array}

In the thousands column we get 9 + 9 = 18. So we bring down 8 and carry the 1 to the ten thousands place.\begin{array}{c} & & &\text{}^1 4 &9 &4 &8 &1 \\ &+ & &4 &9 &1 &0 &0 \\ \hline & & & & 8& 5&8 &1\\ \hline \end{array}

For the ten thousands place we get 1+4+4=9.\begin{array}{c} & & &\text{}^1 4 &9 &4 &8 &1 \\ &+ & &4 &9 &1 &0 &0 \\ \hline & & & 9& 8& 5&8 &1\\ \hline \end{array}

So the answer is:49\,481 + 49\,100 = 98\,581

Example 3

Find the value of 55\,872-2619.

Worked Solution

Create a strategy

Use the standard algorithm method.

Apply the idea

Write it in a vertical algorithm.\begin{array}{c} & & &5 &5 &8 &7 &2 \\ &- & & &2 &6 &1 &9 \\ \hline & \\ \hline \end{array}

In the ones column we can see that 2 is less than 9, so we need to trade 1 ten from the tens place.

So we get 12-9=3 in the ones column and 7 tens becomes 6 tens in the first row.\begin{array}{c} & && 5&5 &8 &6 &\text{}^1 2 \\ &- && &2 &6 &1 &9 \\ \hline & && & & & &3\\ \hline \end{array}

In the tens column, we have 6-1=5:\begin{array}{c} & && 5&5 &8 &6 &\text{}^1 2 \\ &- && &2 &6 &1 &9 \\ \hline & && & & & 5 &3\\ \hline \end{array}

In the hundreds column, we have 8-6=2:\begin{array}{c} & && 5&5 &8 &6 &\text{}^1 2 \\ &- && &2 &6 &1 &9 \\ \hline & && & & 2& 5 &3\\ \hline \end{array}

In the thousands column, we have 5-2=3:\begin{array}{c} & && 5& 5 &8 &6 &\text{}^1 2 \\ &- && &2 &6 &1 &9 \\ \hline & && & 3& 2& 5 &3\\ \hline \end{array}

And for the ten thousands place 5-0=5:\begin{array}{c} & && 5& 5 &8 &6 &\text{}^1 2 \\ &- && &2 &6 &1 &9 \\ \hline & && 5& 3& 2& 5 &3\\ \hline \end{array}

So the answer is:55\,872 - 2619 = 53\,253

Idea summary

A vertical algorithm is very useful when we starting working with more digits. Even though our numbers may be bigger, the problems are solved the same way, just with more steps. For addition, we need to regroup, or trade to the higher place value, but for subtraction we regroup down to the next lower place.

Outcomes

MA3-5NA

selects and applies appropriate strategies for addition and subtraction with counting numbers of any size

MA3-6NA

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

PRACTICE: Addition and subtraction

Ideas