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HotmathMath Homework. Do It Faster, Learn It Better.

Addition is simply finding the sum of two or more numbers, and it's one of the most vital building block in a student's mathematical toolbox. In fact, all of math is just fancy addition! Counting is just adding 1 over and over, multiplying is repeated addition, subtraction is just adding a negative number, and division just asks the question, "How many times did you have to add x to itself to get y?" The list goes on and on!

We add each day without really thinking about it, as addition is used in almost every part of our lives. We use it so often that you may not even think of it as "addition". But learning how to add properly is a key skill that students will need as they advance through their math classes. Let's go through the steps of addition from start to finish.

What is addition of whole numbers?

Addition of whole numbers is simply combining two or more numbers or objects to find a total. The numbers (or objects) are called "terms" or "addends", and the total is called the "sum". Since addition requires you to know your numbers, take the time to review whole numbers before beginning.

At its most basic, addition is an easy process - just count up the total number of objects. Here's an example of how you could use dots to solve the problem $2+7$ :

There are 2 blue dots and 7 red dots. That's a total of 9 dots, so $2+7=9$ .

Addition of whole numbers with carrying over

When you add whole numbers, you'll sometimes need to carry over. We can visualize this by using base-10 blocks. Let's try $115+37$ .

Because 115 is equal to $100+10+5$ , we'll use 1 100-block, 1 10-block, and 5 unit blocks:

37 is equal to $30+7$ , so we can show that with 3 10-blocks and 7 unit blocks:

To add the two numbers, combine all the blocks:

You'll note that we have 12 unit blocks, so we'll trade those (or "carry over") for 1 10-block and 2 unit blocks:

Now we have $100+50+2$ , or 152:

So $115+37=152$ .

When you add by writing out your problem, you'll add each column from right to left. Combining single-unit blocks into a 10-block is called "carrying over", and you can see how it works in the example below:

$\begin{array}{cc}& \hfill 1259\\ +& \hfill \underset{_}{\phantom{\rule{10pt}{0ex}}942}\end{array}$

First, we'll add the ones column together: $9+2$ . Because 11 is too big to fit into the ones, we'll add 1 to the tens column and write 1 as the sum of the ones column:

$\begin{array}{cc}& \hfill {}_{6}\phantom{\rule{12pt}{0ex}}\\ & \hfill 12\overline{)5}\overline{)9}\\ & \hfill \underset{_}{+\phantom{\rule{10pt}{0ex}}9\phantom{\rule{3pt}{0ex}}4\phantom{\rule{5pt}{0ex}}2\phantom{\rule{3pt}{0ex}}}\\ & \hfill 1\phantom{\rule{3pt}{0ex}}\end{array}$

Now, we'll move on to add the tens column. Here, we find that we'll need to carry over as $6+4=10$ . So once again, we'll add 1 to the next column over (in this case, the hundreds column) and write down 0 as the sum of the tens column:

$\begin{array}{cc}& \hfill {}_{3}\phantom{\rule{6pt}{0ex}}{}_{6}\phantom{\rule{12pt}{0ex}}\\ & \hfill 1\overline{)2}\overline{)5}\overline{)9}\\ & \hfill \underset{_}{+\phantom{\rule{10pt}{0ex}}9\phantom{\rule{3pt}{0ex}}4\phantom{\rule{5pt}{0ex}}2\phantom{\rule{3pt}{0ex}}}\\ & \hfill 0\phantom{\rule{6pt}{0ex}}1\phantom{\rule{3pt}{0ex}}\end{array}$

The hundreds column will repeat this process, as $3+9=12$ . We'll write down 2 as the answer to the hundreds column and add 1 to the thousands column:

$\begin{array}{cc}& \hfill {}_{2}\phantom{\rule{6pt}{0ex}}{}_{3}\phantom{\rule{6pt}{0ex}}{}_{6}\phantom{\rule{12pt}{0ex}}\\ & \hfill \overline{)1}\overline{)2}\overline{)5}\overline{)9}\\ & \hfill \underset{_}{+\phantom{\rule{10pt}{0ex}}9\phantom{\rule{5pt}{0ex}}4\phantom{\rule{5pt}{0ex}}2\phantom{\rule{3pt}{0ex}}}\\ & \hfill 2\phantom{\rule{6pt}{0ex}}0\phantom{\rule{5pt}{0ex}}1\phantom{\rule{3pt}{0ex}}\end{array}$

Finally, we'll add up the thousands column. Since there's just one number there, we can simply write it down:

$\begin{array}{cc}& \hfill {}_{2}\phantom{\rule{6pt}{0ex}}{}_{3}\phantom{\rule{6pt}{0ex}}{}_{6}\phantom{\rule{12pt}{0ex}}\\ & \hfill \overline{)1}\overline{)2}\overline{)5}\overline{)9}\\ & \hfill \underset{_}{+\phantom{\rule{10pt}{0ex}}9\phantom{\rule{5pt}{0ex}}4\phantom{\rule{5pt}{0ex}}2\phantom{\rule{3pt}{0ex}}}\\ & \hfill 2\phantom{\rule{4pt}{0ex}}2\phantom{\rule{6pt}{0ex}}0\phantom{\rule{5pt}{0ex}}1\phantom{\rule{3pt}{0ex}}\end{array}$

Addition of whole numbers practice questions

Solve the following problems using whole number addition:

a. $8+2=$

$\begin{array}{cc}& \hfill \overline{)9}\\ & \hfill \underset{_}{+\phantom{\rule{3pt}{0ex}}2\phantom{\rule{3pt}{0ex}}}\\ & \hfill 1\phantom{\rule{5pt}{0ex}}0\phantom{\rule{3pt}{0ex}}\end{array}$

b. $9+10=$

$\begin{array}{cc}& \hfill \overline{)9}\\ & \hfill \underset{_}{+\phantom{\rule{3pt}{0ex}}1\phantom{\rule{3pt}{0ex}}0\phantom{\rule{3pt}{0ex}}}\\ & \hfill 1\phantom{\rule{4pt}{0ex}}9\phantom{\rule{3pt}{0ex}}\end{array}$

c. $64+35=$

$\begin{array}{cc}& \hfill \overline{)6}\overline{)4}\\ & \hfill \underset{_}{+\phantom{\rule{3pt}{0ex}}3\phantom{\rule{4pt}{0ex}}5\phantom{\rule{3pt}{0ex}}}\\ & \hfill 9\phantom{\rule{4pt}{0ex}}9\phantom{\rule{3pt}{0ex}}\end{array}$

d. $242+36=$

$\begin{array}{cc}& \hfill \overline{)2}\overline{)4}\overline{)2}\\ & \hfill \underset{_}{+\phantom{\rule{13pt}{0ex}}3\phantom{\rule{4pt}{0ex}}6\phantom{\rule{3pt}{0ex}}}\\ & \hfill 2\phantom{\rule{4pt}{0ex}}7\phantom{\rule{4pt}{0ex}}8\phantom{\rule{3pt}{0ex}}\end{array}$

e. $7235+431=$

$\begin{array}{cc}& \hfill \overline{)7}\overline{)2}\overline{)3}\overline{)5}\\ & \hfill \underset{_}{+\phantom{\rule{13pt}{0ex}}4\phantom{\rule{6pt}{0ex}}3\phantom{\rule{4pt}{0ex}}1\phantom{\rule{3pt}{0ex}}}\\ & \hfill 7\phantom{\rule{5pt}{0ex}}6\phantom{\rule{6pt}{0ex}}6\phantom{\rule{4pt}{0ex}}6\phantom{\rule{3pt}{0ex}}\end{array}$

Solve the following problems using whole number addition with carrying over:

a. $45+9=$

$\begin{array}{cc}& \hfill {}_{5}\phantom{\rule{12pt}{0ex}}\\ & \hfill \overline{)4}\overline{)5}\\ & \hfill \underset{_}{+\phantom{\rule{10pt}{0ex}}9\phantom{\rule{2pt}{0ex}}}\\ & \hfill 5\phantom{\rule{3pt}{0ex}}4\phantom{\rule{3pt}{0ex}}\end{array}$

b. $56+65=$

$\begin{array}{cc}& \hfill {}_{1}\phantom{\rule{7pt}{0ex}}{}_{6}\phantom{\rule{12pt}{0ex}}\\ & \hfill \overline{)5}\overline{)6}\\ & \hfill \underset{_}{+\phantom{\rule{10pt}{0ex}}6\phantom{\rule{5pt}{0ex}}5\phantom{\rule{2pt}{0ex}}}\\ & \hfill 1\phantom{\rule{5pt}{0ex}}2\phantom{\rule{5pt}{0ex}}1\phantom{\rule{3pt}{0ex}}\end{array}$

c. $235+98=$

$\begin{array}{cc}& \hfill {}_{3}\phantom{\rule{7pt}{0ex}}{}_{4}\phantom{\rule{12pt}{0ex}}\\ & \hfill \overline{)2}\overline{)3}\overline{)5}\\ & \hfill \underset{_}{+\phantom{\rule{10pt}{0ex}}9\phantom{\rule{5pt}{0ex}}8\phantom{\rule{2pt}{0ex}}}\\ & \hfill 3\phantom{\rule{5pt}{0ex}}3\phantom{\rule{5pt}{0ex}}3\phantom{\rule{3pt}{0ex}}\end{array}$

d. $5432+8634=$

$\begin{array}{cc}& \hfill {}_{6}\phantom{\rule{33pt}{0ex}}\\ & \hfill \overline{)5}\overline{)4}\overline{)3}\overline{)2}\\ & \hfill \underset{_}{+\phantom{\rule{5pt}{0ex}}8\phantom{\rule{6pt}{0ex}}6\phantom{\rule{5pt}{0ex}}3\phantom{\rule{5pt}{0ex}}4\phantom{\rule{2pt}{0ex}}}\\ & \hfill 1\phantom{\rule{5pt}{0ex}}4\phantom{\rule{5pt}{0ex}}0\phantom{\rule{5pt}{0ex}}6\phantom{\rule{5pt}{0ex}}6\phantom{\rule{3pt}{0ex}}\end{array}$

e. $36798+2543=$

$\begin{array}{cc}& \hfill {}_{7}\phantom{\rule{4pt}{0ex}}{}_{8}\phantom{\rule{4pt}{0ex}}{}_{10}\phantom{\rule{11pt}{0ex}}\\ & \hfill \overline{)3}\overline{)6}\overline{)7}\overline{)9}\overline{)8}\\ & \hfill \underset{_}{+\phantom{\rule{5pt}{0ex}}2\phantom{\rule{6pt}{0ex}}5\phantom{\rule{5pt}{0ex}}4\phantom{\rule{5pt}{0ex}}3\phantom{\rule{2pt}{0ex}}}\\ & \hfill 3\phantom{\rule{5pt}{0ex}}9\phantom{\rule{5pt}{0ex}}3\phantom{\rule{5pt}{0ex}}4\phantom{\rule{5pt}{0ex}}1\phantom{\rule{3pt}{0ex}}\end{array}$