Using column subtraction to subtract numbers

​​In a nutshell

By arranging a subtraction calculation such that the same place digits are in the same column, you can methodically calculate the answer.

Columns

Numbers in a column formation mean that they are written one on top of the other such that units are in their own column, tens are in their own column, hundreds are in their own columns, and so on. When subtracting, the number being subtracted from goes above the number being subtracted.

Example 1

The following is an example of arranging numbers in columns:

$$\begin {array} { c c c}& 3 & 5 \\- & 1 & 2 \\ \hline& & \\ \hline\end {array}$$​​

Noticed that the units are in one column and the tens are in another column.

​​The method

When in column formation, work from the right and subtract the bottom number from the top number. Write the answer below in the same column. Move on to the left and repeat with the next set of digits.

Example 2

Calculate $$35-12$$.

This can be done with the column method:

$$\begin {array} { c c c}& 3 & 5 \\- & 1 & 2 \\ \hline& & \\ \hline\end {array}$$​​

Starting on the right with the units, calculate $$5-2$$. This gives $$3$$, so write this in between the lines in the same column:

$$\begin {array} { c c c}& 3 & 5 \\- & 1 & 2 \\ \hline& &3 \\ \hline\end {array}$$​​

Now move to the left and work on the tens column. Calculate $$3-1$$. This gives $$2$$, so write this underneath in this column:

$$\begin {array} { c c c}& 3 & 5 \\- & 1 & 2 \\ \hline&2 &3 \\ \hline\end {array}$$​​

Now that you have been through all of the columns, you have the solution. So

$$35-12=\underline{23}$$​​

Borrowing from the left

Example 3

Consider 

$$\begin {array} { c c c}& 5 & 6 \\- & 2 & 9 \\ \hline& & \\ \hline\end {array}$$​​

When you subtract the nine from the six, you will get a negative number. To get around this, "borrow" a one from the column to the left, such that rather than subtracting nine from six, you subtract nine from sixteen. Be aware that since a one has been borrowed from the five in the tens column, this must become four:

$$\begin {array} { c c c}& ^4\cancel{5} & ^16 \\- & 2 & 9 \\ \hline& & \\ \hline\end {array}$$​​

Since $$16-9=7$$, put this seven underneath in the units column:

$$\begin {array} { c c c}& ^4\cancel{5} & ^16 \\- & 2 & 9 \\ \hline& & 7 \\ \hline\end {array}$$​​

Now move to the left to work on the tens column. 

Note: The tens column now has four and two rather than five and two. This is because a one has been borrowed from the five, which turns it into a four.

Since $$4-2=2$$, put a two underneath in the tens column:​

$$\begin {array} { c c c}& ^4\cancel{5} & ^16 \\- & 2 & 9 \\ \hline&2 & 7 \\ \hline\end {array}$$​​

You've worked through all the way to the left, so you have your solution:

$$56-29=\underline{27}$$

​​

Example 4

Calculate $$341-165$$.

Start by setting this up in the column formation:

$$\begin {array} {c c c c}&3& {4} & 1 \\- &1& 6 & 5 \\ \hline& & & \\ \hline\end {array}$$​​

The first step is to work with the units on the right. You want to subtract five from one, but this gives a negative, so you first must borrow a one from the four in the tens column to the left:

​$$\begin {array} {c c c c}&3& ^3\cancel{4} & ^11 \\- &1& 6 & 5 \\ \hline& & & \\ \hline\end {array}$$​

Now in the units column you subtract five from eleven: this gives six. Write this underneath:

​$$\begin {array} {c c c c}&3& ^3\cancel{4} & ^11 \\- &1& 6 & 5 \\ \hline& & &6 \\ \hline\end {array}$$​

Next, you move to the left into the tens column. Here you want to subtract six from three. 

Note: You subtract from three rather than from four since you have borrowed a one from the four for the units column.

Here, $$3-6$$ gives a negative, so again you need to borrow from the left:

$$\begin {array} {c c c c}&^2\cancel3& ^{13}\cancel{4} & ^11 \\- &1& 6 & 5 \\ \hline& & &6 \\ \hline\end {array}$$​​

So now in the tens column you calculate $$13-6$$ to give seven. This goes underneath:

$$\begin {array} {c c c c}&^2\cancel3& ^{13}\cancel{4} & ^11 \\- &1& 6 & 5 \\ \hline& &7 &6 \\ \hline\end {array}$$​​

Finally in the left-most column, subtract the one from the two (this two came from the three, but one was borrowed from it for the tens):

$$\begin {array} {c c c c}&^2\cancel3& ^{13}\cancel{4} & ^11 \\- &1& 6 & 5 \\ \hline& 1 &7 &6 \\ \hline\end {array}$$​​

Hence,

$$341-165=\underline{176}$$​​