Long multiplication

How to multiply two numbers using just a pencil and paper.

The column method

We start with an easy example, and a harder one comes at the end of the article. To begin with, we assume we have two numbers we want to multiply. For now we will work only with natural numbers. Say we want to work out 13 · 72. We write the numbers one under the other and draw a line under them:

$$\underline{\begin{array}{cr}& 13\\ \cdot & 72\end{array}}$$

We start by multiplying the last digit (in the units place) of the bottom number by every digit of the top number. We start from the right again, so first we work out 2 · 3 = 6. We write the result directly under the digit 2. Next we multiply the digit 2 by the second digit from the right, so we work out 2 · 1 = 2. We write this result directly under the 7:

$$\begin{array}{cr}& 13\\ \cdot & 72\\ & 26\end{array}$$

We move one digit to the left in the bottom number, reach the digit 7, and repeat the process. But we have moved up one place (the seven stands for tens, whereas the two stands for units), so we must also move one place along under the line. At the bottom right, under the six, we write a zero. After that we carry on as before.

$$\begin{array}{ccc}& 1& 3\\ \cdot & 7& 2\\ & 2& 6& \\ & & 0\end{array}$$

So we work out 7 · 3 = 21. Here comes the tricky part: we can only write one digit, we cannot write 21 because that is two digits. We solve this by splitting the number into two digits: the right-hand one, 1, goes on the paper, and the left-hand one, 2, we remember for the next round (we carry it). So: we write 1 and carry 2. The one goes right next to the zero.

$$\begin{array}{ccc}& 1& 3\\ \cdot & 7& 2\\ & 2& 6& \\ & 1& 0\end{array}$$

Next we multiply the digit 7 by the digit 1 in the top number: 7 · 1 = 7. Now our carried digit 2 comes into play: we add it to this result, 7 + 2 = 9. We write this number to the left of the digit 1 in the bottom row:

$$\begin{array}{cccc}& & 1& 3\\ \cdot & & 7& 2\\ & & 2& 6& \\ & 9& 1& 0\end{array}$$

There is one last step: adding up the numbers under the line.

$$\begin{array}{ccccc}& & 2& 6& \\ +& 9& 1& 0\\ & 9& 3& 6\end{array}$$

We get 936, which is exactly the product 13 · 72.

Why it works

We can split a multiplication into simpler products and sums like this: 13 · 72 is the same as 13 · 2 + 13 · 70. You may see taking out a common factor in this, but plain common sense is enough: if I add up thirteen seventy-two times, it is the same as adding thirteen seventy times and then adding two more thirteens.

The method does nothing more than produce these partial products. If you look, the numbers we got under the line were 26 and 910. And indeed 26 = 2 · 13 and 910 = 70 · 13.

For this to work out, we had to write a zero in the second row, because what we were really working out was 7 · 13. To turn this into the product 70 · 13, we need to put one more zero on the end of the result.

A harder example

Work out 28 · 617. First we write the numbers one under the other:

$$\underline{\begin{array}{cccc}& & 2& 8\\ \cdot & 6& 1& 7\end{array}}$$

In the first step we work out 7 · 8 = 56. We write 6 and carry 5.

$$\begin{array}{cccc}& & 2& 8\\ \cdot & 6& 1& 7\\ & & & 6\end{array}$$

Next, 7 · 2 = 14, and we add the 5 we carried: 14 + 5 = 19. We write 9 and carry 1.

$$\begin{array}{cccc}& & 2& 8\\ \cdot & 6& 1& 7\\ & & 9& 6\end{array}$$

Since there is nothing left to multiply, we write the carried one:

$$\begin{array}{cccc}& & 2& 8\\ \cdot & 6& 1& 7\\ & 1& 9& 6\end{array}$$

On to the second row. We write a zero and multiply: 1 · 8 = 8, so we write 8:

$$\begin{array}{cccc}& & 2& 8\\ \cdot & 6& 1& 7\\ & 1& 9& 6\\ & & 8& 0\end{array}$$

Next, 1 · 2 = 2, so we write 2:

$$\begin{array}{cccc}& & 2& 8\\ \cdot & 6& 1& 7\\ & 1& 9& 6\\ & 2& 8& 0\end{array}$$

And on to the third row. We move another place along, so we write one more zero than before:

$$\begin{array}{cccc}& & 2& 8\\ \cdot & 6& 1& 7\\ & 1& 9& 6\\ & 2& 8& 0\\ & & 0& 0\end{array}$$

And we multiply: 6 · 8 = 48, we write 8 and carry 4.

$$\begin{array}{cccc}& & 2& 8\\ \cdot & 6& 1& 7\\ & 1& 9& 6\\ & 2& 8& 0\\ & 8& 0& 0\end{array}$$

Next, 6 · 2 = 12, we add 4, so we get 12 + 4 = 16. We write 6 and carry 1.

$$\begin{array}{cccc}& & 2& 8\\ \cdot & 6& 1& 7\\ & 1& 9& 6\\ & 2& 8& 0\\ 6& 8& 0& 0\end{array}$$

Since there is nothing left to multiply, we write the carried one:

$$\begin{array}{ccccc}& & & 2& 8\\ \cdot & & 6& 1& 7\\ & & 1& 9& 6\\ & & 2& 8& 0\\ 1& 6& 8& 0& 0\end{array}$$

And now we add up all three numbers under the line:

$$\begin{array}{cccccc}& & & 1& 9& 6\\ & & & 2& 8& 0\\ +& 1& 6& 8& 0& 0\\ & 1& 7& 2& 7& 6\end{array}$$