47

__x 8__

__That would be tough to pull off in your head. However, in school, you learn that you can simply do it in two steps.__

1. 7 x 8 is 56. Write the 6 and carry the five.

5

47

__x 8__

6

2. 4 x 8 is 32. Add the five to that and you get 37.

5

47

376

And there you go. This works great for small problems like 2x1, 3x1, 4x1, 5x1, or even ones like 2x2 or 3x2. However, what about something like a 4x4. Your paper would look like:

8922

47

__x 8__376

And there you go. This works great for small problems like 2x1, 3x1, 4x1, 5x1, or even ones like 2x2 or 3x2. However, what about something like a 4x4. Your paper would look like:

8922

__x 6137__
62454

267660

892200

__+ 53532000__

54754314

This method is basically the distributive property, where you are multiplying each number by every other number. For instance, a 3x3 algebraically would look like:

Let x = the first digit of the first factor

Let y = the second digit of the first factor

Let z = the third digit of the first factor

Let a = the first digit of the second factor

Let b = the second digit of the second factor

Let c = the third digit of the second factor

Let P = the product of the two factors

P = (z + 10y + 100x)(c + 10b + 100a)

Using this traditional multiplication method, this product then looks like:

P = c(z + 10y + 100x) + 10b(z + 10y + 100x) + 100a(z + 10y + 100x)

This looks pretty complicated. What this method then does is takes it one step further by distributing the a, b, and c through their parentheses. The one thing that it leaves factored out is the powers of ten that are coefficients of the a, b, and c. You don't multiply through the 10 and 100 that are in front of the b and the a.

P = [(c)(z) + (c)(10y) + (c)(100x)] + 10[(b)(z) + (b)(10y) + (b)(100x)] + 100[(a)(z) + (a)(10y) + a(100x)]

You might be thinking that you don't do that much work in your head. I mean, you've been doing this since third grade! But think through what you are doing. How are you creating the terms to add up?

The first term is created by multiplying the last digit of the second factor, or c, by each digit of the first factor. There is a multiplication method called partial products which would multiply the c by the 10y and 100x, but the traditional method uses the power of zero rule to just write the (c)(10y) to the left of the (c)(z), and the (c)(100x) to the left of that.

The second term is created similarly by multiplying the second digit of the second factor, which we denoted as b, by each digit of the first factor. However, we must remember to put a zero down first, which was something constantly driven into our heads by our math teachers. This is taking into account the ten in front of the brackets. The power of zero rule states that when multiplying by ten, we can just tack a zero onto the end of the other factor. So, we tack a zero onto the end of this number that was found by multiplying b by every digit of the first factor.

The third term is also created by multiplying the first digit of the second factor, which we called a, by each digit of the first factor. Since we have a 100 outside of the brackets, we must tack two zeros onto the end of this number.

Finally, we add up these terms, which is exactly what the equation does above to get the product.

Is this really how we want to get to our product? Can't we combine the terms in a little bit of an easier way?

Let's try it. Let's fully distribute every single term through. We will get:

P = 1cz + 10cy + 100cx + 10bz + 100by + 1000bx + 100az + 1000ay + 10000ax

How about we treat the coefficients as if they are variables. In that case, we have lots of like terms to add.

P = 1(cz) + 10(cy + bz) + 100(cx + by + az) + 1000(bx + ay) + 10000(ax)

That looks fairly simple, or at least compared to the first method. However, if you think it through, you might suspect that the numbers you are multiplying together will not be in natural places in your actual problem.

I want you to try to test this. Write the following on a piece of scrap paper lying around.

xyz

__x abc__

Now, the first step is just cz. Draw a erasable line from z to c. You can see the relationship between the first two numbers being multiplied. Now, erase it.

Here's where it may appear to get awkward. (cy + bz). Draw an erasable line from c to y, and from b to z. What is a easy relationship here?

In fact, you have just made a cross. These two completely random terms have formed a cross on your paper. You can erase the cross now.

Try the third terms, which there are a lot of them! Draw lines between c and x, b and y, and a and z. Believe it or not, you will get a three-way cross, which might resemble an asterisk slightly.

If you continue this process with the other two, you will find just another cross on the left, and then a line on the left side of your problem.

Because of these crosses, this method is called the "Criss-Cross Method." With this crossing pattern in mind, let's try an example. How about 143 x 822.

143

__x 822__

If you'll remember, we first have a line on the right side of the problem, connecting the 3 and the 2. So, we must multiply three and two, which will give us six. We then place that directly under the 3 and the 2.

143

__x 822__

6

Now, we must draw our first cross, which is also on the right side of the problem. We have a line connecting the 4 and the 2, and a line connecting the 3 and the 2. Remember that it resembles an X, it is not two parallel lines connecting the four to the left two and the three to the right two. It is the other way around.

Let's find the products of the connected numbers. 4 x 2 = 8 and 3 x 2 = 6. If you'll remember from our algebra, we must add together these products. 8 + 6 = 14.

We cannot stick a 14 there, just like we can't do in any other multiplication method. We will have to put the four, and carry the one.

143

__x 822__1

46

Now, we have our three-way cross. We must do three computations: 1 x 2, 4 x 2, and 3 x 8. Don't try to do them all at once, just do one at a time. 1 x 2 = 2, and add to that the one we carried to get three. 4 x 2 = 8, and add that to the three to get eleven. 3 x 8 = 24, and add that to the eleven to get thirty-five. We can then put the five and carry the three.

Notice that after each computation, we only had to remember one additional number. This is the power of keeping a running total after each step. The whole purpose of this is to make your final result easier and less messy, which keeping a running total will do.

143

__x 822__3

546

Now, we will do the next criss-cross, which is the 1 x 2 and the 4 x 8. 1 x 2 = 2, which we can add to the three we carried to get five. 4 x 8 = 32, which we can add to that five to get thirty-seven. We will put the seven, and carry the three.

143

__x 822__3

7546

Now, we are in home stretch. 1 x 8 = 8, which we will add to the three to get eleven. Since there is nothing to carry over to, we will just put the eleven there.

143

__x 822__3

117546

Believe it or not, we have just found our answer. We can also continue this up to a 4x4, or a 5x5, or even higher if you want. The only change is the number of criss-crosses you make. For a 4x4, you do your first line, then a cross, then a three-way cross on the right side of the problem, then a four-way cross incorporating all of the numbers in the problem, followed by a three-way cross on the left side of the problem, followed by a two-way cross, followed by that same last line.

This may look a little cumbersome, especially with the gigantic crosses you have to do. However, this method is a lot easier than your traditional multiplication once you start getting into larger factors, and will take much less time to master.

Additionally, it is fun to impress your friends with. If you have them give you a 3x3 multiplication problem, you will be able to knock it out in ten seconds or less, and will not have written a single number on your paper. Since I am known for my mental math, the criss-cross method is something I can use to fool people into thinking I handled the whole problem in my head in one shot, rather than writing down each digit after I compute it.