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3.06 Addition with regrouping

Lesson

Are you ready?

Do you remember how to solve addition problems by writing them down the page,  using an algorithm  ?

Examples

Example 1

Find the value of 41 + 56.

Worked Solution
Create a strategy

Use the addition algorithm.

Apply the idea

Write the addition in a vertical algorithm.\begin{array}{c} & &4 &1 \\ &+ &5 &6 \\ \hline & \\ \hline \end{array}

Add the smallest place value first. So 1 + 6 = 7. \begin{array}{c} & &4 &1 \\ &+ &5 &6 \\ \hline & & &7 \\ \hline \end{array}

Then add the next place value. 4 + 5 = 9. \begin{array}{c} & &4 &1 \\ &+ &5 &6 \\ \hline & &9 &7 \\ \hline \end{array}

So 41 + 56 = 97.

Idea summary

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 with regrouping

Sometimes when we add numbers together we need to regroup, or use trading. Let's see when we need to do this, and how we do it.

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Examples

Example 2

Find the value of 19 + 25.

Worked Solution
Create a strategy

Use the addition algorithm.

Apply the idea

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

Begin with the units place value. So 9 + 5 = 14.

Place 4 under the units column in the answer section and carry forward 1 to the tens column. \begin{array}{c} & &\text{}^1 1 &9 \\ &+ &2 &5 \\ \hline & & &4 \\ \hline \end{array}

Add the numbers in the tens column. Since 1 has been carried over from the units place, we need to add 1 + 1 + 2 = 4. \begin{array}{c} & &\text{}^1 1 &9 \\ &+ &2 &5 \\ \hline & &4 &4 \\ \hline \end{array}

So 19 + 25 = 44.

Idea summary

The largest value a digit can be is 9. If we add and end up with more than 9, we need to trade, or regroup, to the next digit to the left.

Outcomes

MA2-5NA

uses mental and written strategies for addition and subtraction involving two-, three-, four and five-digit numbers

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