Notes Don't bother trying to divide any number by zero—the solution doesn't exist. In fact, the very statement of the problem is invalid. In this lesson, you will not be required to divide a whole number by one that has a larger value. In other words, you will not see a problem such as 8 ÷ 16. This is a valid problem, but the quotient is a fraction, and fractions do not belong to the whole-number system. - Any value divided by one is equal to the original value.
| Example: 5 ÷ 1 = 5 | - Zero divided by any whole number (except 0) is zero.
| Example: 0 ÷ 2 = 0 | |
Sometimes you will find division problems expressed in the vertical form ...
... and quite often in the division box form:
Examples and Exercises
Dividing Small Whole Numbers Use these interactive examples and exercises to strengthen your understanding and build your skills: | |
Division With Remainders
Most combinations of whole numbers do not divide evenly. In other words, their quotient cannot be expressed as a simple whole number. - 2 divides evenly into 6: 2 ) 6 = 3
- 7 does not divide evenly into 37: 7 ) 37 = ?
7 divides into 37 five times ... with a remainder of 2 
|
Procedure When numbers do not divide evenly: - Find the largest number of times the divisor will divide into the dividend. This is the quotient.
- To determine the remainder, multiply the quotient by the divisor, then subtract the result from the dividend.
|
Sometimes it is necessary to divide a number into a smaller value, such as:
2 ÷ 7 6 ÷ 9 0 ÷ 2
In the whole number system, remainders come to the rescue.
When the divisor is larger than the dividend, the quotient is zero with a remainder that is equal to the divisor.
2 ÷ 7 = 0 R 7 6 ÷ 9 = 0 R 9 0 ÷ 2 = 0 R 2
Note You can always check the result of a division operation by multiplying the divisor by the quotient. Your answer is correct if the divisor times the quotient is equal to the dividend. If there is a remainder, you should add it to the result. Examples Problem: 30 ÷ 5 = 6 Check: 5 x 6 = 30 Problem 51 ÷ 4 = 12 R 3 Check: 4 x 12 = 48 48 + 3 = 51 |
Examples and Exercises
Dividing With Remainders Use these interactive examples and exercises to strengthen your understanding and build your skills: | |
Using Short Division
Short division is a convenient way to solve division problems when the divisor is a fairly small number. Here are some examples of problems where short division works well:
21
4 ) 84 | 7
6 ) 42 | 4
12 ) 48 | 2133
2 ) 4266 | 4
23 ) 92 |
Example 1
Problem 2 ) 684 = _____ | |
| Procedure | |
- Divide 2 into the 6, and write the result above the 6
| |
- Divide 2 into the 8, and write the result above the 8
| |
- Divide 2 into the 4, and write the result above the 4
| |
Solution 2 ) 684 = 342 | |
Examples and Exercises
Simple Short Division Use these interactive examples and exercises to strengthen your understanding and build your skills: | |
Example 2
Problem 4 ) 736 = _____ | |
| Procedure | |
- Divide 4 into the 7: 1 R 3
Write the 1 over the 7 and "carry" the remainder to the next number in the dividend | |
- Divide 4 into the 33: 8 R 1
Write the 8 over the 33 and "carry" the remainder to the next number in the dividend. | |
- Divide 4 into the 16.
Write the result over the 16. | |
Solution 4 ) 736 = 184 | |
Examples and Exercises
Short Division With "Carrying" Use these interactive examples and exercises to strengthen your understanding and build your skills: | |
In all the examples of short division you have been working so far, the first digit in the dividend is larger than the divisor. More often, however, the first digit in the dividend is smaller than the divisor. When this happens, you must try dividing the divisor into the first two digits in the dividend.
Example 3
Divide 5 into 360
?
5 ) 360 | 5 does not divide into 3 at least one time. |
7
5 ) 36 10 | So begin by dividing 5 into 35. |
7 2
5 ) 36 10 | Then complete the division. |
Examples and Exercises
Short Division With Larger Divisor Use these interactive examples and exercises to strengthen your understanding and build your skills: | |
Also, you will find that most division problems end up with a remainder.
Example 4
Problem 5 ) 823 = _____ | |
| Procedure | |
- Divide 5 into the 8: 1 R 3
Write the 1 over the 8 and "carry" the remainder to the next number in the dividend | |
- Divide 5 into the 32: 6 R 2
Write the 6 over the 32 and "carry" the remainder to the next number in the dividend. | |
- Divide 5 into the 23.
Write the result over the 23, and show the remainder.. | |
Solution 5 ) 823 = 164 R 3 | |
Example 5
Problem 6 ) 458 = _____ | |
| Procedure | |
- Try to divide the 6 into 4, but it does not divide at least one time. So divide the 6 into 45: 7 R 3
Write the 7 over the 45, and "carry" the remainder to the next number in the dividend | |
- Divide 6 into the 38: 6 R 2
Write the 6 over the 38, and show the remainder of 2. | |
Solution 6 ) 458 = 76 R 2 | |
Examples and Exercises
Short Division With Remainders It is very important that you continue working these examples and exercises until you can consistently work them without making any errors. | |
Using Long Division
| You should use long division when the divisor has more than one digit. In the first example, let's use a problem that is easily solved with short division. We'll use long division, however, to show how it works. | Note It is possible to use short division for problems having more than one digit in the divisor. However, the "carrying " operations can become very cumbersome. | |
Example 6
Problem 2 ) 78 | |
- Divide the 2 into the 7.
- Place the result over the 7
- Place the product of 2 x 3 under the 7
| |
- Subtract
| |
- "Bring down the next digit, the 8
| |
- Divide the 2 into the 18.
- Place the result over the 8 in the dividend.
- Place the product of 2 x 9 under the 18
| |
- Subtract
The problem is done when there are no more numbers to "bring down" from the divisor. | |
Solution 2 ) 78 = 39 | |
Here is an example that is a bit more complicated and has a remainder term.
Example 7
Problem 14 ) 878 = ____ | |
- Divide 14 into 87
- Place the result over the 87
- Place the product of 14 x 6 under the 87
- Subtract
| |
- Bring down the 8
| |
- Divide 14 into 38
- Place the result over the 8
- Place the product of 14 x 2 under the 38
- Subtract
| 62
14 ) 878
–840
38
–28
10 | | |
- Get the remainder
The problem is done, because the 14 (the divisor) cannot be divided into 10 (result of the last subtraction). The 10 becomes the remainder of the division operation. | 62
14 ) 878
–840
38
–28
10 | R 10 | |
Solution 14 ) 878 = 62 R 10 | |
Examples of Long Division - Part 1
Carefully study the details of these examples of long division. Do not quit until you are sure you understand every step in every example you see.
Exercises for Long Division - Part 1
Use long division to solve these examples. Keep working them until you have completely mastered the procedure.
Here are some more examples and exercises. They are a bit more complicated than the previous set, but this is exactly the level of work you are expected to do.
Examples of Long Division - Part 2
Carefully study the details of these examples of long division. Do not quit until you are sure you understand every step in every example you see.
Exercises for Long Division - Part 2
Use long division to solve these examples. Keep working them until you have completely mastered the process.
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