SECTION 1-WHOLE NUMBER ARITHMETIC

MULTIPLICATION-PART 1

I am going to start our study of multiplication by looking at the addition problem "5 + 5 = 10." Here you have two 5's which add up to a total of 10. You could think of this as having two groups of 5. With that in mind you could say "two groups of five is equal to 10."

Now let's take a look at three 5's.
5 + 5 + 5 = 15." Here you have three 5's which add up to a total of 15. You could think of this as having three groups of 5. With that in mind you could say "three groups of five is equal to 15."

Now let's take a look at four 5's.
5 + 5 + 5 + 5 = 20." Here you have four 5's which add up to a total of 20. You could think of this as having four groups of 5. With that in mind you could say "four groups of five is equal to 20."

What you have just seen is the very principle behind multiplication. In multiplication you have a certain number of groups (or sets), and each of the groups (or each of the sets) is the same size. The total number of items you have can be obtained by multiplying the number of groups by the size of each group. The answer of any multiplication problem is called the product. Now as you probably surmised this product is equivalent to adding the group size to itself however many times required by the number of groups. The above problems regarding the number five could be stated as the following multiplication problems:

5 multiplied by 2 is 10
5 multiplied by 3 is 15
5 multiplied by 4 is 20

Rather than continually write the words "multiplied by" we will use mathematics symbols for multiplication. Now there are three different ways to write a multiplication problem. In fact, if I wanted to multiply 5 by 2 I could write it any of the following ways

5 x 2
5 * 2
5 · 2

I will be using the "x" symbol to designate multiplication.

With this in mind, the three sentences listed above (5 multiplied by 2, 3, and 4) could be written as follows:

5 multiplied by 2 can be written as "5 x 2"
5 multiplied by 3 can be written as "5 x 3"
5 multiplied by 4 can be written as "5 x 4"

Here are a few more multiplication problems

"Three being multiplied by two is equal to six" could be written as "3 x 2 = 6"
You can think of this problem as having 2 sets, and each set has 3 elements. Altogether you would have "3 + 3" or 6 elements.
"Two being multiplied by three is equal to six" could be written as "2 x 3 = 6"
You can think of this problem as having 3 sets, and each set has 2 elements. Altogether you would have "2 + 2 + 2" or 6 elements.
"Six being multiplied by four is equal to twenty-four" could be written as "6 x 4 = 24"
You can think of this problem as having 4 sets, and each set has 6 elements. Altogether you would have "6 + 6 + 6 + 6" or 24 elements.
"Four being multiplied by six is equal to twenty-four" could be written as "4 x 6 = 24"
You can think of this problem as having 6 sets, and each set has 4 elements. Altogether you would have "4 + 4 + 4 + 4 + 4 + 4" or 24 elements.

Notice that 3 x 2 = 2 x 3 and 4 x 6 = 6 x 4.

Before you go on be forewarned that I assume that you can multiply single digit numbers. If you need help multiplying single digit numbers then go to the appendix.

Multiplication is a little more involved than addition or subtraction, but you can handle it!

First of all the number that is to be multiplied is called the multiplicand, and the number that you are multiplying by is called the multiplier. The solution to the problem is called the product. So if I was going to multiply 6 by 3, then 6 would be the multiplicand, and 3 would be the multiplier. Since 6 times 3 is 18, the product is 18.

I'll start out by showing you how to multiply by a a single digit. Remember to take it one step at a time, and remember that we are going from right to left.

Multiply the digit on the right hand side of the multiplicand by the multiplier. If the product that you arrive at is less than 10, then record this number. If the product is larger than 10, then sepearte it into the required number of ones, and the required number of tens. Record the "ones-value" and carry the "tens-value" onto the next column to the left.
Now multiply the next digit in the multiplicand by the multiplier. If necessary add this product to the number you carried over from the previous column. If the result that you arrive at is less than 10, then record this number. If the result is larger than 10, then sepearte it into the required number of ones, and the required number of tens. Record the "ones-value" and carry the "tens-value" onto the next column to the left.
Continue to repeat the previous step until you reach the final digit of the multiplicand. (i.e. the digit furthest to the left).
When you get to the last column of the multiplicand, you need not seperate the "ones-value" and "tens-value." Simply record the number received after multiplying that digit by the multiplier and adding any number carried over from the previous column.

This is illustrated in the examples below.

Multiplication example 1

324 x 8

Here 324 is the multiplicand and 8 is the multiplier.
We are interested in 8 groups of 324 (i.,e. the result of adding 324 to itself a total of 8 times).
Our first step is to line the problem as shown below
                       324
                     x   8
We start with the digit on the right hand side of the multiplicand. In this case that digit is 4.
8 x 4 = 32
32 = 30 + 2 = 3-ten's + 2-one's.
Write down the 2, and carry the 3 (Yes, here we go with "carrying" again)

3 324 x 8 2

The next digit in the multiplicand is 2. Now multiply 8 by 2, and then add the carried 3 to this product.
8 x 2 = 16 and 16 + 3 = 19
19 = 10 + 9 = 1-ten + 9-ones
Write down the 9, and carry the 1

1 324 x 8 92

We have now reached the last digit in the multiplicand: 3. Multiply 3 by 8, and then add our carried 1 to this product.
8 x 3 = 24 and 24 + 1 = 25.
Since we have reached the end of the multiplicand, record 25 in the finished product.

324 x 8 2592

324 x 8 = 2,592

Multiplication example 2

6238 x 2

Here 6238 is the multiplicand and 2 is the multiplier.
We are interested in 2 groups of 6238 (i.,e. the result of adding 6238 to itself a total of 2 times).
Our first step is to line the problem as shown below
                       6238
                     x    2
We start with the right-hand digit of the multiplicand. In this case that digit is 8.
8 x 2 = 16
16 = 10 + 6 = 1-ten + 6-ones.
Write down the 6 and carry the 1

1 6238 x 2 6

Moving on to the next digit of the multiplicand we have the number 3. Multiply 3 by 2, and then add the carried 1 to this product.
3 x 2 = 6 and 6 + 1 = 7
Write down the 7

6238 x 2 76

The next digit in 6238 is 2.
2 x 2 = 4

6238 x 2 476

Now we have reached the last column.
2 x 6 = 12

6238 x 2 12476

6238 x 2 = 12,476

Next I'll be multiplying a number with at least two digits by another number with at least two digits. But first, it's time for you to practice multiplying.
Click on the "generate a multiplication problem" button to see your exercise. Place your answer in the text box and then click the "check my answer" button. The computer will then tell you whether or not your answer is correct.

After you complete the problem, you can generate another problem by first clicking the "reset button" and then clicking the "generate a multiplication problem" button. Continue as many times as you wish, and then:

Text, images, & source code c2003 Martin Selditch