Dividing Whole Numbers

So far we have explored addition, subtraction, and multiplication. Now let’s consider division. Suppose you have the [latex]12[/latex] cookies and want to package them in bags with [latex]4[/latex] cookies in each bag. How many bags would we need?

You might put [latex]4[/latex] cookies in first bag, [latex]4[/latex] in the second bag, and so on until you run out of cookies. Doing it this way, you would fill [latex]3[/latex] bags.

In other words, starting with the [latex]12[/latex] cookies, you would take away, or subtract, [latex]4[/latex] cookies at a time.

Division is a way to represent repeated subtraction just as multiplication represents repeated addition. Instead of subtracting [latex]4[/latex] repeatedly, we can write

We read this as twelve divided by four and the result is the quotient of [latex]12[/latex] and [latex]4[/latex]. The quotient is [latex]3[/latex] because we can subtract [latex]4[/latex] from [latex]12[/latex] exactly [latex]3[/latex] times. We call the number being divided the dividend and the number dividing it the divisor. In this case, the dividend is [latex]12[/latex] and the divisor is [latex]4[/latex]. In the past you may have used the notation [latex]4\overline{)12}[/latex] , but this division also can be written as [latex]12\div 4, 12\text{/}4, \frac{12}{4}[/latex]. In each case the [latex]12[/latex] is the dividend and the [latex]4[/latex] is the divisor.

Division is performed on two numbers at a time. When translating from math notation to English words, or English words to math notation, look for the words of and and to identify the numbers.

When the divisor or the dividend has more than one digit, it is usually easier to use the [latex]4\overline{)12}[/latex] notation. This process is called long division.

Let’s work through the process by dividing [latex]78[/latex] by [latex]3[/latex].

Divide the first digit of the dividend, [latex]7[/latex], by the divisor, [latex]3[/latex].
The divisor [latex]3[/latex] can go into [latex]7[/latex] two times since [latex]2\times 3=6[/latex] . Write the [latex]2[/latex] above the [latex]7[/latex] in the quotient.
Multiply the [latex]2[/latex] in the quotient by [latex]2[/latex] and write the product, [latex]6[/latex], under the[latex]7[/latex].
Subtract that product from the first digit in the dividend. Subtract [latex]7 - 6[/latex] . Write the difference, 1, under the first digit in the dividend.
Bring down the next digit of the dividend. Bring down the [latex]8[/latex].
Divide [latex]18[/latex] by the divisor, [latex]3[/latex]. The divisor [latex]3[/latex] goes into [latex]18[/latex] six times.
Write [latex]6[/latex] in the quotient above the [latex]8[/latex].
Multiply the [latex]6[/latex] in the quotient by the divisor and write the product, [latex]18[/latex], under the dividend. Subtract [latex]18[/latex] from [latex]18[/latex].

 

We would repeat the process until there are no more digits in the dividend to bring down. In this problem, there are no more digits to bring down, so the division is finished.

So [latex]78\div 3=26[/latex]

 

Check by multiplying the quotient times the divisor to get the dividend. Multiply [latex]26\times 3[/latex] to make sure that product equals the dividend, [latex]78[/latex].

[latex]\begin{array}{c}\hfill \stackrel{1}{2}6\\ \hfill \underset{\text{___}}{\times 3}\\ \hfill 78 \end{array}[/latex]

It does, so our answer is correct. [latex]\checkmark[/latex]

When dividing whole numbers we have to keep a couple of properties in mind.

You may be asked to divide even larger numbers, don’t panic, the process is the same no matter how big the numbers get.

Divide the following and check your answer by multiplying:

1. [latex]2,596\div 4[/latex]
2. [latex]4,506\div 6[/latex]
3. [latex]7,263\div 9[/latex]

Show Solution

1.

Let’s rewrite the problem to set it up for long division.
Divide the first digit of the dividend, [latex]2[/latex], by the divisor, [latex]4[/latex].
Since [latex]4[/latex] does not go into [latex]2[/latex], we use the first two digits of the dividend and divide [latex]25[/latex] by [latex]4[/latex]. The divisor [latex]4[/latex] goes into [latex]25[/latex] six times.
We write the [latex]6[/latex] in the quotient above the [latex]5[/latex].
Multiply the [latex]6[/latex]in the quotient by the divisor [latex]4[/latex] and write the product, [latex]24[/latex], under the first two digits in the dividend.
Subtract that product from the first two digits in the dividend. Subtract [latex]25 - 24[/latex] . Write the difference, [latex]1[/latex], under the second digit in the dividend.
Now bring down the [latex]9[/latex] and repeat these steps. There are [latex]4[/latex] fours in [latex]19[/latex]. Write the [latex]4[/latex] over the [latex]9[/latex]. Multiply the [latex]4[/latex] by [latex]4[/latex] and subtract this product from [latex]19[/latex].
Bring down the [latex]6[/latex] and repeat these steps. There are [latex]9[/latex] fours in [latex]36[/latex]. Write the [latex]9[/latex] over the [latex]6[/latex]. Multiply the [latex]9[/latex] by [latex]4[/latex] and subtract this product from [latex]36[/latex].
So [latex]2,596\div 4=649[/latex] .
Check by multiplying.
It equals the dividend, so our answer is correct.

 

2.

Let’s rewrite the problem to set it up for long division.
First we try to divide [latex]6[/latex] into [latex]4[/latex].
Since that won’t work, we try [latex]6[/latex] into [latex]45[/latex]. There are [latex]7[/latex] sixes in [latex]45[/latex]. We write the [latex]7[/latex] over the [latex]5[/latex].
Multiply the [latex]7[/latex] by [latex]6[/latex] and subtract this product from [latex]45[/latex].
Now bring down the [latex]0[/latex] and repeat these steps. There are [latex]5[/latex] sixes in [latex]30[/latex]. Write the [latex]5[/latex] over the [latex]0[/latex]. Multiply the [latex]5[/latex] by [latex]6[/latex] and subtract this product from [latex]30[/latex].
Now bring down the [latex]6[/latex] and repeat these steps. There is [latex]1[/latex] six in [latex]6[/latex]. Write the [latex]1[/latex] over the [latex]6[/latex]. Multiply [latex]1[/latex] by [latex]6[/latex] and subtract this product from [latex]6[/latex]
Check by multiplying.
It equals the dividend, so our answer is correct.

 

3.

Let’s rewrite the problem to set it up for long division.
First we try to divide [latex]9[/latex] into [latex]7[/latex].
Since that won’t work, we try [latex]9[/latex] into [latex]72[/latex]. There are [latex]8[/latex] nines in [latex]72[/latex]. We write the [latex]8[/latex] over the [latex]2[/latex].
Multiply the [latex]8[/latex] by [latex]9[/latex] and subtract this product from [latex]72[/latex].
Now bring down the [latex]6[/latex] and repeat these steps. There are [latex]0[/latex] nines in [latex]6[/latex]. Write the [latex]0[/latex] over the [latex]6[/latex]. Multiply the [latex]0[/latex] by [latex]9[/latex] and subtract this product from [latex]6[/latex].
Now bring down the [latex]3[/latex] and repeat these steps. There are [latex]7[/latex] nines in [latex]63[/latex]. Write the [latex]7[/latex] over the [latex]3[/latex]. Multiply the [latex]7[/latex] by [latex]9[/latex] and subtract this product from [latex]63[/latex].
Check by multiplying.
It equals the dividend, so our answer is correct.

So far all the division problems have worked out evenly.

For example, if we had [latex]24[/latex] cookies and wanted to make bags of [latex]8[/latex] cookies, we would have [latex]3[/latex] bags.

But what if there were [latex]28[/latex] cookies and we wanted to make bags of [latex]8[/latex]?

Start with the [latex]28[/latex] cookies.

 

Try to put the cookies in groups of eight.

 

There are [latex]3[/latex] groups of eight cookies, and [latex]4[/latex] cookies left over. We call the [latex]4[/latex] cookies that are left over the remainder and show it by writing R4 next to the [latex]3[/latex]. (The R stands for remainder.) To check this division we multiply [latex]3[/latex] times [latex]8[/latex] to get [latex]24[/latex], and then add the remainder of [latex]4[/latex].

[latex]\begin{array}{c}\hfill 3\\ \hfill \underset{\text{___}}{\times 8}\\ \hfill 24\\ \hfill \underset{\text{___}}{+4}\\ \hfill 28\end{array}[/latex]

Divide the following and check your answer by multiplying:

1. [latex]1,439\div 4[/latex]
2. [latex]1,461\div 13[/latex]

Show Solution

1.

Let’s rewrite the problem to set it up for long division.
First we try to divide [latex]4[/latex] into [latex]1[/latex]. Since that won’t work, we try [latex]4[/latex] into [latex]14[/latex]. There are [latex]3[/latex] fours in [latex]14[/latex]. We write the [latex]3[/latex] over the [latex]4[/latex].
Multiply the [latex]3[/latex] by [latex]4[/latex] and subtract this product from [latex]14[/latex].
Now bring down the [latex]3[/latex] and repeat these steps. There are [latex]5[/latex] fours in [latex]23[/latex]. Write the [latex]5[/latex] over the [latex]3[/latex]. Multiply the [latex]5[/latex] by [latex]4[/latex] and subtract this product from [latex]23[/latex].
Now bring down the [latex]9[/latex] and repeat these steps. There are [latex]9[/latex] fours in [latex]39[/latex]. Write the [latex]9[/latex] over the [latex]9[/latex]. Multiply the [latex]9[/latex] by [latex]4[/latex] and subtract this product from [latex]39[/latex]. There are no more numbers to bring down, so we are done. The remainder is [latex]3[/latex].
Check by multiplying.
So [latex]1,439\div 4[/latex] is [latex]359[/latex] with a remainder of [latex]3[/latex]. Our answer is correct.

 

2.

Let’s rewrite the problem to set it up for long division.	[latex]13\overline{)1,461}[/latex]
First we try to divide [latex]13[/latex] into [latex]1[/latex]. Since that won’t work, we try [latex]13[/latex] into [latex]14[/latex]. There is [latex]1[/latex] thirteen in [latex]14[/latex]. We write the [latex]1[/latex] over the [latex]4[/latex].
Multiply the [latex]1[/latex] by [latex]13[/latex] and subtract this product from [latex]14[/latex].
Now bring down the [latex]6[/latex] and repeat these steps. There is [latex]1[/latex] thirteen in [latex]16[/latex]. Write the [latex]1[/latex] over the [latex]6[/latex]. Multiply the [latex]1[/latex] by [latex]13[/latex] and subtract this product from [latex]16[/latex].
Now bring down the [latex]1[/latex] and repeat these steps. There are [latex]2[/latex] thirteens in [latex]31[/latex]. Write the [latex]2[/latex] over the [latex]1[/latex]. Multiply the [latex]2[/latex] by [latex]13[/latex] and subtract this product from [latex]31[/latex]. There are no more numbers to bring down, so we are done. The remainder is [latex]5[/latex]. [latex]1,462\div 13[/latex] is [latex]112[/latex] with a remainder of [latex]5[/latex].
Check by multiplying.
Our answer is correct.

Sometimes it might not be obvious how many times the divisor goes into digits of the dividend. We will have to guess and check numbers to find the greatest number that goes into the digits without exceeding them.

Divide the following. Check your answer by multiplying.

[latex]74,521\div 241[/latex]

Show Solution

Let’s rewrite the problem to set it up for long division.	[latex]241\overline{)74,521}[/latex]
First we try to divide [latex]241[/latex] into [latex]7[/latex]. Since that won’t work, we try [latex]241[/latex] into [latex]74[/latex]. That still won’t work, so we try [latex]241[/latex] into[latex]745[/latex]. Since [latex]2[/latex] divides into [latex]7[/latex] three times, we try [latex]3[/latex]. Since [latex]3\times 241=723[/latex] , we write the [latex]3[/latex] over the [latex]5[/latex] in [latex]745[/latex]. Note that [latex]4[/latex] would be too large because [latex]4\times 241=964[/latex] , which is greater than [latex]745[/latex].
Multiply the [latex]3[/latex] by [latex]241[/latex] and subtract this product from [latex]745[/latex].
Now bring down the [latex]2[/latex] and repeat these steps. [latex]241[/latex] does not divide into [latex]222[/latex]. We write a [latex]0[/latex] over the [latex]2[/latex] as a placeholder and then continue.
Now bring down the [latex]1[/latex] and repeat these steps. Try [latex]9[/latex]. Since [latex]9\times 241=2,169[/latex] , we write the [latex]9[/latex] over the [latex]1[/latex]. Multiply the [latex]9[/latex] by [latex]241[/latex] and subtract this product from [latex]2,221[/latex].
There are no more numbers to bring down, so we are finished. The remainder is [latex]52[/latex]. So [latex]74,521\div 241[/latex] is [latex]309[/latex] with a remainder of [latex]52[/latex].
Check by multiplying.

Divide Whole Numbers in Applications

We will use the same strategy we used in previous sections to solve applications. First, we determine what we are looking for. Then we write a phrase that gives the information to find it. We then translate the phrase into math notation and simplify it to get the answer. Finally, we write a sentence to answer the question.