Whole number arithmetic

Free Web Hosting Provider - Web Hosting - E-commerce - High Speed Internet - Free Web Page

How many twos do you need to reach 14?
How many fives are there in 15?

The answers to the above questions are 7 and 3 respectively.
Seven sets of two would equal 14 (i.e. 7 x 2 = 14)
Three sets of five would equal 15 (i.e. 3 x 5 = 15)

The above two questions are specific examples of division. In division you are asking the follwing question:
"I have a set (or a group) of a certain size. How many of these sets (or groups) will it take to reach a given number?"
The answer to this question is the quantity that we want to reach, divided by the size of each set (or group).

With this in mind we could write the questions posed in the beginning of this section as follows:

Now rather than continually write the words "divided by" we will can use any of the following shorthand symbols.

Before you go on be forewarned that I assume that you can use the multiplication table for both multiplication and division. If you have any trouble with this, or if you need more help learning the definition of division, then please see the appendix.

Before beginning our study of long division let's look at one more relationship regarding division.

First let's take a look at the division problems "15 divided by 5" and "15 divided by 3."
15/5 = 3. Here the dividend is 15, the divisor is 5 and the quotient is 3.
15/3 = 5. Here the dividend is 15, the divisor is 3 and the quotient is 5.
All I did was switch the divisor and the quotient. Why does this work?

Remember when I told you that in multiplication it doesn't matter which number you choose as the multiplicand, and which numbers you choose as the multiplier? In this example 3 x 5 = 5 x 3. Or stated in different manner "five groups of three" is the same total as "three groups of five."

Now how does that relate to division? Well remember that division is the opposite of multiplication. Therefore the statement "15 / 3 = 5" is telling us that "you need 5 groups of three to equal 15." And the statement "15 / 5 = 3" is telling us that "you need 3 groups of 5 to equal 15." However "5 groups of three" is the same total as "3 groups of 5." Therefore these two division problems boil down to equivalent multiplication problems.

Let's look at that one more time.

18/6 = 3. Here the dividend is 18, the divisor is 6 and the quotient is 3.
18/3 = 6. Here the dividend is 18, the divisor is 3 and the quotient is 6.
Again I simply switched the divisor and the quotient. Why does this work?

Remember it doesn't matter which number you choose as the multiplicand, and which numbers you choose as the multiplier. In this example 6 x 3 = 3 x 6. Or stated in different manner "three groups of six" is the same total as "six groups of three."

Now how does that relate to these division problems? Well remember that division is the opposite of multiplication. Therefore the statement "18 / 3 = 6" is telling us that "you need 6 groups of three to equal 18." And the statement "18 / 6 = 3" is telling us that "you need 3 groups of 6 to equal 18." However "3 groups of 6" is the same total as "6 groups of 3." Therefore these two division problems boil down to equivalent multiplication problems.

This relationship of switching the divisor and quotient will work for any division problem.

Now let's start our study of long division. Just as we did in multiplication we will begin our study with one digit divisors.
Unlike multiplication, we will be reading the dividend from left to right rather than right to left.

The best way I know of to teach division is to dive right in. So I will show you six examples in this section. After that you will have a chance to practice your division skills.

72 ¸ 2
The dividend is 72 and the divisor is 2.

	First set up the problem as shown on the left.
	Now we will look at the first digit of the dividend. In this case that number is 7. What is the maximum number of times that the divisor go into this number? In other words what is the maximum number of times that 2 can go into 7? What do I mean by this? What number when multiplied by 2 will yield a product very close to 7, without going over 7? Well let's see. For our first guess let's try 1. 2 x 1 = 2 which is less than 7, but we can get closer to 7 than that. For our second guess let's try 2. 2 x 2 = 4 which is less than 7, but we can get closer to 7 than that. For our next guess let's try 4. 2 x 4 = 8 which is greater than 7. Since 4 was too large and 2 was too small, let's try 3. 2 x 3 = 6 which is less than 7. Therefore the maximum number of times that 2 can go into 7 is 3. Write the three in the area reserved for the quotient, and place it directly above the 7 that appears in 72. At one time in this section I had the number 2 (the product of 1 and our divisor 2) and I knew that I could get closer to 7 than that. At another time I had the number 4 (the product of 2 and our divisor 2) and I knew that I could get closer to 7 than that. How did I know this? Well the general rule is as follows: "Multiply your guess by the divisor, and then subtract this product from the number in question. The result must be less than the divisor." My first guess was 1. 1 x 2 = 2 and 7 - 2 = 5 However 5 is larger than the divisor, therefore I knew that I could get closer to 7 than that. My second guess was 2. 2 x 2 = 4, and 7 - 4 = 3 However 3 is larger than the divisor, therefore I knew that I could get closer to 7 than that. My last guess was 3. 2 x 3 = 6 , and 7 - 6 = 1. The number 1 is less than the divisor. Therefore 3 was the correct guess.
	Now multiply the number you just received (3) by the divisor (2), and place this product directly under the number you just investigated (7). In other words multiply 2 by 3 and place the resulting product (6) directly under the 7. Now subtract the 6 from the 7 giving you a difference of 1. This is your remainder.
	Now look at the next digit of the dividend. Place that digit on the right hand side of the remainder. In this case our remainder is 1 and the next digit of the dividend is 2. Placing the 2 on the right hand side of the remainder will give us the number 12.
	Now will will investigate 12. What is the maximum number of times that the divisor go into this number? In other words what is the maximum number of times that 2 can go into 12? What do I mean by this? What number when multiplied by 2 will yield a product very close to 12, without going over 12? Well let's see. For our first guess let's try 4. 2 x 4 = 8, and 12 - 8 = 4. That difference is larger than our divisor, which indicates that we can get closer to 12 than 8. For our next guess let's try 5. 2 x 5 = 10, and 12 - 10 = 2. That difference is equal to our divisor. The rule that I stated back in step 2 specifically said that the difference must be less than the divisor. The number 2 is not less than itself. Therefore we can get closer to 12 than 10. For our next guess let's try 6. 2 x 6 = 12. We couldn't possibly get closer to 12 than this! Therefore the maximum number of times that 2 can go into 12 is 6. Write the six in the area reserved for the quotient, and place it directly above the 2 that appears in 72.
	Now multiply the number you just received (6) by the divisor (2), and place this product directly under the number you just investigated (12). In other words multiply 2 by 6 and place the resulting product (12) directly under the 12. Now subtract these numbers. This will give you a difference of 0. There is no remainder. There are no more digits in the dividend so we are now finished.

72 ¸ 2 = 36
What does this result mean?
Well the division problem "72 ¸ 2" is asking us "how many groups of 2 are required to reach 72?"
The quotient of 32 is telling us that "we need 36 groups of 2 in order to reach 72."
So if we wanted to check our solution then we would multiply 36 and 2. The result of this multiplication should be 72.
Assignment: verify that "36 x 2 = 72"

318 / 6
The dividend is 318 and the divisor is 6.

	First set up the problem as shown on the left.
	Now we will look at the first digit of the dividend (remember that in division we are reading left to right). In this case that number is 3. What is the maximum number of times that the divisor go into this number? In other words what is the maximum number of times that 6 can go into 3? What do I mean by this? What number when multiplied by 6 will yield a product very close to 3, without going over 3? In this case the answer is zero. Even 1 multiplied by 6 will yield a result larger than 3. So what do we do now? Look at the first two digits of the dividend. The first digit of the dividend taken together with the second digit of the dividend forms the number 31. Since 6 cannot go into 3, maybe it can go into 31. What is the maximum number of times that 6 can go into 31? What do I mean by this? What number when multiplied by 6 will yield a product very close to 31, without going over 31? Let's see. For our first guess let's try 4. 6 x 4 = 24, and 31 - 24 = 7. That difference is larger than our divisor, therefore we can get closer to 31 than 24. For our next guess let's try 6. 6 x 6 = 36, which is larger than 31. Therefore 6 was too large. Since 6 was too large, and 4 was too small, let's try 5. 6 x 5 = 30 and 31 - 30 = 1. That difference is less than our divisor, which indicates that we arrived at the correct number. Therefore the maximum number of times that 6 can go into 31 is 5. Write the five in the area reserved for the quotient, and place it directly above the 1 that appears in 318.
	Now multiply the number you just received (5) by the divisor (6) and place this product directly under the number you just investigated (31). In other words multiply 5 by 6 and place the resulting product (30) directly under the 31. Now subtract the 30 from the 31 giving you a difference of 1. This is your remainder.
	Now look at the next digit of the dividend. Place that digit on the right hand side of the remainder. In this case our remainder is 1 and the next digit of the dividend is 8. Placing the 8 on the right hand side of the 1 will give us the number 18.
	Now will will investigate 18. What is the maximum number of times that the divisor go into this number? In other words what is the maximum number of times that 6 can go into 18? What do I mean by this? What number when multiplied by 6 will yield a product very close to 18, without going over 18? Well let's see. Let's try the number 1. 6 x 2 = 12 and 18 - 12 = 6. This difference is not less than our divisor. Therefore we can get closer to 18 than 12. For our next guess let's try the number 3. 6 x 3 = 18. We couldn't possibly get closer to 18 than this! Therefore the maximum number of times that 6 can go into 18 is 3. Write the three in the area reserved for the quotient, and place it directly above the 8 that appears in 318.
	Now multiply the number you just received (3) by the divisor (6), and place this product directly under the number you just investigated (18). In other words multiply 3 by 6 and place the resulting product (18) directly under the 18. Now subtract these numbers. This will give you a difference of 0. There is no remainder. There are no more digits in the dividend so we are now finished.

318 / 6 = 53
What does this result mean?
Well the division problem "318 /6 " is asking us "how many groups of 6 are required to reach 318?"
The quotient of 53 is telling us that "we need 53 groups of 6 in order to reach 318."
So if we wanted to check our solution then we would multiply 53 and 6. The result of this multiplication should be 318.
Assignment: verify that "53 x 6 = 318"

Are you following me so far. Good. These next two examples will show that nothing sometimes plays an important role in division. In other words the importance of zero.

848 / 8
The dividend is 848 and the divisor is 8.

	First set up the problem as shown on the left.
	Now we will look at the first digit of the dividend (remember that in division we are reading left to right). In this case that number is 8. What is the maximum number of times that the divisor go into this number? In other words what is the maximum number of times that 8 can go into 8? The answer is 1. Now you have three tasks to perform. First write a one in the area reserved for the quotient, and place it directly above the first 8 that appears in 848. Next, multiply this 1 by the divsor (8) getting a product of 8. Place this product directly below the 8 that we were just discussing. Finally, subtract 8 from 8 getting 0. There is no remainder.
	Now look at the next digit of the dividend. In this case that digit is 4. Place that digit on the right hand side of our zero remainder. You will get "04." Now we all know that "04" is the same as 4, but we will keep the zero just so everything lines up nice and neat. In this case zero is a place holder.
	What is the maximum number of times that our divisor (8) will go into 4? In other words what number when multiplied by 8 will yield a product very close to 4, without going over 4? The answer is zero. Even 1 multiplied by 8 would yield a product larger than 4. Since 8 can go into 4 a maximum of zero times, write a zero in the area reserved for the quotient, and place it directly above the 4 that appears in 848. Now multiply this 0 by the divisor (8) yielding a product of zero. Place this zero below the "04" and subtract. You will end up with a remainder of 4. Notice that subtracting zero from 4 did not alter it at all.
	Now look at the next digit of the dividend. In this case that digit is 8. Place that 8 on the right hand side of our remainder 4. You will end up with the number 48. Here you see the net result of placing a zero in the quotient. Back in step 3 we dropped the 4 from the dividend creating the number "04." Our divisor couldn't go into 4, so we placed a zero in the quotient. Now we dropped the next digit from the divident creating the number "048."
	What is the maximum number of times that our divisor (8) will go into 48? In other words what number when multiplied by 8 will yield a product very close to 48, without going over 48? Well let's see. For our first guess let's try 5. 8 x 5 = 40 and 48 - 40 = 8. Since 8 is not less than our divisor, we know that we can get closer to 48 than 40. For our next guess let's try 7. 8 x 7 = 56. This is larger than 48, which indicates that our guess of 7 was too large. Since 5 was too small anbd 7 was too large, let's try the number 6. 8 x 6 = 48. We couldn't possibly get any closer to 48 than that! Therefore the maximum number of times that 8 goes into 48 is 6. Write a six in the area reserved for the quotient, and place it directly above the second 8 which appears in 848. Now multiply this 6 by the divisor (8) yielding a product of 48. Place this product directly below the 48 and subtract. You will end up with a remainder of 0. Since there are no more digits in the dividend, then we are finished.

848 / 8 = 106
What does this result mean?
Well the division problem "848 /8" is asking us "how many groups of 8 are required to reach 848?"
The quotient of 106 is telling us that "we need 106 groups of 8 in order to reach 848."
So if we wanted to check our solution then we would multiply 106 and 8. The result of this multiplication should be 848.
Assignment: verify that "106 x 8 = 848"

10405 ¸ 5
The dividend is 10405 and the divisor is 5.

	First set up the problem as shown on the left.
	Now we will look at the first digit of the dividend (remember that in division we are reading left to right). In this case that number is 1. Our divisor (5) cannot go into 1 even once. So we should look at the first two digits of the dividend as though they formed a two-digit number. By doing this we are now investigating the number 10. What is the maximum number of times that the divisor 5 can go into 10? Well let's see. 5 x 1 = 5 and 10 - 5 = 5. This difference is not less than our divisor which indicates that we can get closer to 10 than 5. 5 x 2 = 10. We couldn't possible get closer to 10 than that! Write a two in the area reserved for the quotient, and place it directly above the first zero that appears in 10405. Now multiply this 2 by the divisor (5) getting a product of 10. Place this product directly below the "10" that we just investigated. Now subtract this product from that "10" getting a remainder of 0.
	Now look at the next digit of the dividend. In this case that digit is 4. Place that digit on the right hand side of our zero remainder. You will get "04." Now we all know that "04" is the same as 4, but we will keep the zero just so everything lines up nice and neat. In this case zero is a place holder.
	What is the maximum number of times that our divisor (5) will go into 4? In other words what number when multiplied by 5 will yield a product very close to 4, without going over 4? The answer is zero. Even 1 multiplied by 5 would yield a product larger than 4. Write a zero in the area reserved for the quotient, and place it directly above the 4 that appears in 10405. Ordinarily we would multiply this zero by the divisor getting a product of zero. We would then place this product below the "04." We would then subtract this zero from 4 getting 4. And finally we would drop the next digit from the divisor thereby changing the "04" into "040." A very similar process as this was illustrated in the previous example.
	What is the maximum number of times that our divisor (5) will go into 40? In other words what number when multiplied by 5 will yield a product very close to 40, without going over 40? Well let's see. For our first guss let's try 7. 5 x 6 = 35 and 40 - 35 = 5. The number 10 is not smaller than than our divisor, so we can get closer to 40 than 35. Let's try the number 8. 5 x 8 = 40. We couldn't possibly et any closer to 40 than that. Therefore 5 goes into 40 a maximum of 8 times. Write eight in the area reserved for the quotient, and place it directly above the second zero that appears in 10405. Now multiply this 8 by the divisor and place that product (in this case 40) directly below the "40" that we were just investigating. Now subtract the two numbers leaving you with a remainder of 0.
	Now look at the next digit of the dividend. In this case that digit is 5. Place that digit on the right hand side of our zero remainder. You will get "05." Now we all know that "05" is the same as 5, but we will keep the zero just so everything lines up nice and neat. In this case zero is a place holder.
	What is the maximum number of times that our divisor (5) will go into 5? In other words what number when multiplied by 5 will yield a product very close to 5, without going over 5? The answer is 1. So write the one in the area reserved for the quotient, and place it direcly above the 5 than appears in 10405. Now multiply this 1 by the divisor and place that product (in this case 5) directly below the "05" that we were just investigating. Now subtract the two numbers leaving you with a remainder of 0. There are no more numbers in the dividend. Therefore we are finished.

10405 ¸ 5 = 2081
What does this result mean?
Well the division problem "10405 ¸ 5" is asking us "how many groups of 5 are required to reach 10405?"
The quotient of 2081 is telling us that "we need 2081 groups of 5 in order to reach 10405."
So if we wanted to check our solution then we would multiply 2081 and 5. The result of this multiplication should be 10405.
Assignment: verify that "2081 x 5 = 10405"

So far everything worked out nice and neat. In the end there were no remainders. But in life everything is not always so neat. This will be illustrated for you in these next two examples.

313 ¸ 6
The dividend is 313 and the divisor is 6.

	First set up the problem as shown on the left.
	Now we will start by looking at the first digit of the dividend (remember that in division we are reading left to right). In this case that number is 3. Our divisor (6) cannot go into 3 even once. So we should look at the first two digits of the dividend as though they formed a two-digit number. By doing this we are now investigating the number 31. What is the maximum number of times that the divisor 6 can go into this 31? The answer turns out to be 5 (Assignment: confirm this yourself). Write the 5 in the area reserved for the quotient, and place it directly above the 1 that appears in 313. Now multiply the divisor 6 by this answer getting a product of 30. Place this product directly below the "31" that we just investigated. Now subtract the "30" from that "31" getting a remainder of 1.
	Now look at the next digit of the dividend. In this case that digit is 3. Place that digit on the right hand side of our one remainder. You will get "13." Now we will examine the number 13.
	What is the maximum number of times that the divisor 6 can go into this 13? The answer turns out to be 2 (Assignment: confirm this yourself). Write the 3 in the area reserved for the quotient, and place it directly above the second 3 that appears in 313. Now multiply the divisor 6 by this answer getting a product of 12. Place this product directly below the "13" that we just investigated. Now subtract the "12" from that "13" getting a remainder of 1. There are no more numbers in the dividend. Therefore we are finished.

313 ¸ 6 = 52 with a remainder of 1.
Some math books write this solution as "52r1."
What does this result mean?
Well let's look at the division problem "313 ¸ 6"
It basically is saying that we have a group of 6 objects, and asking us how many of these groups are required to reach 313 objects?
The quotient of "52r1" is telling us that "we need 52 group of 6 to get near 313, and then we need to add 1 more object." In other words if we multiplied 52 by 6, and then added 1 to this product the result should be 313.
52 x 6 = 312 and 312 + 1 = 313.
Therefore our solution has been verified.

90 ¸ 7
The dividend is 90 and the divisor is 7.

	First set up the problem as shown on the left.
	Now we will start by looking at the first digit of the dividend (remember that in division we are reading left to right). In this case that number is 9. What is the maximum number of times that the divisor 7 can go into 9? The answer turns out to be 1 (Assignment: confirm this yourself). Write the 1 in the area reserved for the quotient, and place it directly above the 9 that appears in 90. Now multiply the divisor 7 by this answer getting a product of 7. Place this product directly below the "9" that we just investigated. Now subtract the "7" from that "9" getting a remainder of 2.
	Now look at the next digit of the dividend. In this case that digit is 0. Place that digit on the right hand side of our two remainder. You will get "20." Now we will examine the number 20.
	What is the maximum number of times that the divisor 7 can go into this 20? The answer turns out to be 2 (Assignment: confirm this yourself). Write the 2 in the area reserved for the quotient, and place it directly above the 0 that appears in 90. Now multiply the divisor 7 by this answer getting a product of 14. Place this product directly below the "20" that we just investigated. Now subtract the "14" from that "20" getting a remainder of 6. There are no more numbers in the dividend. Therefore we are finished.

90 ¸ 7 = 12 with a remainder of 6.
Some math books write this solution as "12r6."
What does this result mean?
Well let's look at the division problem "90 ¸ 7"
It basically is saying that we have a group of 7 objects, and asking us how many of these groups are required to reach 90 objects?
The quotient of "12r6" is telling us that "we need 12 group of 7 to get near 90, and then we need to add 6 more objects." In other words if we multiplied 12 by 7, and then added 6 to this product the result should be 90.
12 x 7 = 84 and 84 + 6 = 90.
Therefore our solution has been verified.

Next we will discuss divisors larger than ten. But first, it's time for you to practice division.

Click on the "generate a division 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 division problem" button. Continue as many times as you wish, and then:

SECTION 1-WHOLE NUMBER ARITHMETIC