The mathematical expression 3 5 represents three groups with five items in each group. To find the product, students can build a model of three groups with five items in each group as shown below.
Students can also use repeated addition to find the product. They can add 3 five times: 3 + 3 + 3 + 3 + 3 = 15.
Remember that multiplication undoes division and division undoes multiplication. In other words, since 3
5 = 15, then 15
5 = 3. Since division and multiplication are inverse operations, students can use models similar to the models used in multiplication, to divide. In the expression 15
3, you begin with fifteen items and want to know how many groups you can make with three items in each group. The answer, or quotient, is the number of groups.
3 = 5
Since multiplication is a form of repeated addition, division is a form of repeated subtraction. For example, 15
3 asks you to repeatedly subtract 3 from 15 until you reach zero: c15 - 3 = 12 - 3 = 9 - 3 = 6 - 3 = 3 - 3 = 0. This process required 3 to be subtracted 5 consecutive times, so again we see that 15
3 = 5. The number that is divided is called the dividend and the number which the dividend is being divided by is the divisor. The answer to a division problem is the quotient.
divisor = quotient
As students master their basic division facts, the need will arise for students to learn how to divide larger dividends. It would be prudent to begin with 2-digit dividends by 1-digit divisor to
introduce the long division algorithm. Although students may know the quotient for the problem, the need to carefully learn the division algorithm will allow students to quickly move to larger numbers. Look at the division problem below.
It is sometimes very helpful if you think of division as multiplication. The above division problem can be written as 306
6 =. or.
6 = 306. In the multiplication problem.
6 = 306, 306 is the product, 6 is a factor and. is a missing factor. In the division problem 306
6 =. 306 can be thought of as the product, 6 as a factor, and. as the missing factor. The missing factor is really the answer, or quotient, of a division problem. So, when looking at
, we are trying to find what number multiplied by 6 will give us 306. Understanding this concept will significantly help students learn the long division algorithm.
When beginning the long division algorithm, realize that you are asking questions such as "what number times 6 is less than or equal to 300?" Remember that the 3 is in the hundreds place. Since the answer is 5 tens, or 50, a 5 can be written above the 0 in the tens place as shown below. Since 5 tens = 50, and 50
6 = 300, we can take 300 away from 306 leaving 6. The process is repeated when asking "what number times 6 is less than or equal to 6." Since the answer is 1, a 1 is written above the 6 in the ones place. Since 6
1 is 6, 6 is taken away from 6 leaving 0. The problem is complete since the amount left over, the remainder, is less than the divisor 6. The complete problem is shown below.