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### Lattice Multiplication Explained

```
Date: 10/20/1999 at 09:55:54
From: Julie Durham a
Subject: Lattice Multiplication

I know how to figure out the Lattice Multiplication procedure, but I
don't understand why it works. We learned the traditional way of
multiplying, and it works with this but I don't see how it works when
you put the numbers on a square. Why and how does it work? I asked my
teacher and she said, "Magic."

I was never really great at math but I like this and I want to
understand why it works. How many algorithms for multiplication are
there in this world? We only learned it one way and now I know two and
I like this one better than the other way. Can you help, please? Can
you give me other examples of multiplication algorithms?
```

```
Date: 10/20/1999 at 12:18:09
From: Doctor Peterson
Subject: Re: Lattice Multiplication

Hi, Julie.

I'm glad you find lattice multiplication useful. In case you want to
see more of it, here's a page in our archives on it:

Lattice Multiplication
http://mathforum.org/dr.math/problems/susan.8.340.96.html

You might also find Napier's Rods (or "bones") interesting:

Napier's Rods
http://mathforum.org/dr.math/problems/phyllis03.05.99.html

There were many methods of multiplication in earlier times (say the
16th and 17th centuries), but they were generally similar to either
our usual method or the lattice method. You might want to check a good
book on the history of math to see how they worked, but I think only

As for why the method works, let's try inventing the method ourselves,
and see why we'd do it the way we do. Math is never magic - it's just
that the reasons are sometimes pretty well hidden. We'll use this
example, as in the archived answer I referred you to:

469
x  37
-----

The basis of any method of multiplying is the distributive property:

a x (b + c) = a x b + a x c

In this case,

469 x 37 = (400 + 60 + 9) x (30 + 7)
= (400 + 60 + 9) x 30 + (400 + 60 + 9) x 7
= 400 x 30 + 60 x 30 + 9 x 30 + 400 x 7 + 60 x 7 + 9 x 7

In other words, we can break each number up into a sum of terms, one
term for each digit, and the product will be the sum of all possible
products of a term from one number and a term from the other. How can
we do that easily? A multiplication table does the same thing - a
table of all products. So let's make a multiplication table for these
terms:

400     60     9
+-------+------+-----+
|       |      |     |
| 12000 | 1800 | 270 | 30
|       |      |     |
+-------+------+-----+
|       |      |     |
|  2800 |  420 |  63 | 7
|       |      |     |
+-------+------+-----+

(I've written the labels for the rows on the right rather than the
left, just because I know that they would be in the way on the left.)

Notice that the products along a diagonal from top right to bottom
left have the same number of zeroes (the same exponent of 10). That's
because when you go left a column and down a row, you add a zero and
take it back off. So we can add the numbers along each diagonal line,
then sum the results:

400     60     9
+-------+------+-----+
|       |      |     |
| 12000 | 1800 | 270 | 30
|       |      |     |
+-------+------+-----+
/ |       |      |     |
/   | 2800  |  420 |  63 | 7
/     |       |      |     |
/       +-------+------+-----+
/       /       /      /
12000 +  4600  +  690  +  63 = 17353

We can drop the zeroes and just write the product of the non-zero
digits, and put the zeroes back in when we add:

4      6      9
+------+------+------+
|      |      |      |
|  12  |  18  |  27  | 3
|      |      |      |
+------+------+------+
/ |      |      |      |
/   |  28  |  42  |  63  | 7
/     |      |      |      |
/       +------+------+------+
/       /      /      /
12      46     69     63

12
46
69
63
-----
17353

But we can make it a little easier by noticing that the first digit of
each product will add into the second digit of the diagonal to its
left. So we can split each product into two digits and draw diagonals
to show this:

4      6      9
+------+------+------+
| 1   /| 1   /| 2   /|
|   /  |   /  |   /  | 3
| /  2 | /  8 | /  7 |
+------+------+------+
/ | 2   /| 4   /| 6  / |
/   |   /  |   /  |   /  | 7
/     | /  8 | /  2 | /  3 |
/       +------+------+------+
/       /      /      /      /
1      5      22     15      3

1
5
22
15
3
-----
17353

Now if you just do the carries as you write the sum of each diagonal,
you'll have the lattice method, with as little writing as possible:

4      6      9
+------+------+------+
| 1   /| 1   /| 2   /|
|   /  |   /  |   /  | 3
| /  2 | /  8 | /  7 |
+------+------+------+
/ | 2   /| 4   /| 6  / |
/   |   /  |   /  |   /  | 7
/     | /  8 | /  2 | /  3 |
/       +------+------+------+
/       /      /      /      /
1      7       3      5      3

You can see the same process at work in our normal method, which we

469
x  37
-----
63  <-- 7x9
42   <-- 7x60
28    <-- 7x400
27   <-- 30x9
18    <-- 30x60
12     <-- 30x400
-----
17353  <-- (30+7) x (400+60+9)

(notice the same six products we had before), or write more compactly
as:

469
x  37
-----
3283  <-- 7x(400+60+9)
1407   <-- 30x(400+60+9)
-----
17353  <-- (30+7)x(400+60+9)

Have fun using the lattice!

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Elementary Multiplication

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