Addition (with regrouping, 3+ digits)

Once you have learned to add with regrouping, you can add numbers of any size. The process that you use with multi-digit numbers is the same as the process used with two-digit numbers, you just keep moving over, place value by place value.

Essentiallly, every time you add a column of digits and get a two-digit answer, you place the ones digit of your answer in the answer row below the added column, and you "carry" or regroup the tens digit to the next column of numbers.  Whenever you get to the last column of digits that you need to add, you write all digits of the sum in the answer row. 

The process of "carrying" is explained in the lesson Addition (with regrouping, 2 digits), and the theory behind the process still holds with multiple digits.  Whenever you add a column of digits, the tens digit in your answer is the equalivalent to a digit in the column to the left, so you can just add it into the next column.

Example: 

$9367+2456=$

To use the "regrouping" or "carrying" process, you always want to start by setting up your problem vertically, lining up the place values:

$$\begin{array}{r} &9367\\+\!\!\!\!\!\!&2456\\ \hline \end{array}$$

Then we would point out that the column the farthest to the right is the ones column, and the next column is the 10s column.

So, first we add the ones colum:

$$\begin{array}{r} &93\overset{1}{6}\mathbf{7}\\+\!\!\!\!\!\!&245\mathbf{6}\\ \hline &\mathbf{3} \end{array}$$

We get 13, so we put the 3 from 11 in the ones column and we bring the 10 from the 11 up to the ten's column.

Then we add the items in the 10s column (including the carried 1):

$$\begin{array}{r} &9\overset{1}{3}\mathbf{\overset{1}{6}}7\\+\!\!\!\!\!\!&24\mathbf{5}6\\ \hline &\mathbf{2}3 \end{array}$$

We get 12, so we put the 2 from 12 in the ones column and we bring the 10 from the 12 up to the next column in the problem, the hundreds column.

Then we add the items in the 100s column (including the carried 1):

$$\begin{array}{r} &9\mathbf{\overset{1}{3}}\overset{1}{6}7\\+\!\!\!\!\!\!&2\mathbf{4}56\\ \hline &\mathbf{8}23 \end{array}$$

We get 8.  This is a one digit answer, so we put it in the answer row and move on to the next column, the thousands column.

Then we add the items in the 1000s column, and because this is the last column, we'll put the full sum of this answer in the answer row (no carrying):

$$\begin{array}{r} &\mathbf{9}\overset{1}{3}\overset{1}{6}7\\+\!\!\!\!\!\!&\mathbf{2}456\\ \hline &\mathbf{11}823 \end{array}$$

We get 11 and put both digits in the answer row.  

$9367+2456=11823$

Overall, we want students to understand that we write numbers vertically and line them up, so that numbers with the same value are in the same column (ones with ones, tens with tens).

Then we want them to understand that each column holds only one digit.  When we have a number that spans two digits (11, 31, 56), then we carry the front number (in this case, the ten) to the next column.

From here, you just practice.  Make sure that they always remember to write out their carry numbers, and check for accuracy.  Even if students don't fully understand what they are doing, as they get better at the skill and gain more math knowledge, everything that they are doing will start to make more sense.  By explaining the concept behind the algorithm, you give the students the tools that they need to do math through understanding, rather than just through rote memory (even if, in the early stages, they rely a lot on memory too!).