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Multiplication on paper

Kapitoly: Multiplication, Multiplication on paper, Multiplying negative numbers

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

How to multiply on paper

We will start with an easy example, a more complex example is at the end of the article. At the beginning, we assume we have two numbers that we want to multiply. We will work only with natural numbers for now. Suppose we want to multiply the numbers 13 · 72. We'll write these numbers down and add a line underneath them:

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

Start multiplying the last digit (in place of the ones) in the bottom number with all the digits in the top number. We start again from the right, so we multiply 2 · 3 = 6 first . We write the result directly under the digit 2. Next, we multiply the digit 2 with the second digit from the right, so we multiply 2 · 1 = 2. We write this result directly under 7:

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

Moving one digit to the left in the bottom number, we get to the digit 7, and do the same procedure. However, since we have moved an order of magnitude (the seven represents tens, while the two represents ones), we have to move an order of magnitude below the line as well. On the bottom right, below the six, we write a zero. Then we continue in the same way.

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

So we multiply by 7 · 3 = 21. This is where the betrayal comes in - we can only add one digit, we can't write 21 - that's two digits. We solve this by splitting the number into two digits: we write the right one, 1, on the paper and remember the left one, 2, until the next round. So: 1 we write on paper, 2 we remember. We write it next to the zero.

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

Next, we multiply the digit 7 with the digit 1 in the top number: 7 · 1 = 7. Now comes our digit 2, which we have memorized: we add it to this result: 7 + 2 = 9. We write this number to the left of the digit 1 in the bottommost line:

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

That leaves the last step: add up the numbers below the line.

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

13 · 72The next step is to finish the line, which is the last step: we need to add the last line to the bottom line.

Why this works

We can decompose multiplication into simpler sums and addends like this: we can write that 13 · 72 is the same as 13 · 2 + 13 · 70. You can see the reproach in this, but simple reason is enough: if I add seventy times the number thirteen, it is the same as if I added seventy times the number thirteen and then added two more thirteens.

The procedure described produces nothing but these successive sums. If you'll notice, the footnotes give us the numbers 26 and 910. Yet 26 = 2 · 13 and 910 = 70 · 13.

To make it really fit, we had to add a zero in the second line - because we were actually counting 7 · 13. To get the product of 70 · 13, we need to add another zero to the result.

A more complicated example

Calculate: 28 · 617. The first thing we do is write the numbers underneath:

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

In the first step, we calculate: 7 · 8 = 56. We write 6, 5 goes next.

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

Next: 7 · 2 = 14, we add 5 from last time: 14 + 5 = 19. Let's type 9, 1 next.

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

Since there's nothing left to multiply, we add one:

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

We go to the second line. We add a zero and multiply: 1 · 8 = 8, we type 8:

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

Next: 1 · 2 = 2, write 2:

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

And we go to the third line. We move up one more line, so we write one more zero than before:

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

And we multiply: 6 · 8 = 48, 8 we write, 4 goes on.

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

Next: 6 · 2 = 12, we add 4, so we get 12 + 4 = 16. 6 we write, 1 goes on.

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

Since there's nothing left to multiply, we add one:

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

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

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