Box models are powerful because they provide a general method for dividing large numbers. And, as we've seen, they can also handle cases with remainders.

Let's consider the following three-digit by one-digit division problem.

630 \div 3

We set up our box model in the usual way.

3

6 3 0

We now follow our usual four steps:

Step 1: Pick a (fairly large) multiple of 3 that is easy to compute but is no larger than 630. Let's pick 3 \times {\color{blue}200} = 600.

We write {\color{blue}200} on top of the box and subtract 3 \times{\color{blue}200} from 630 inside the box.

\color{blue}200
3

6 3 0
-\! 6 0 0
\color{red}3 \color{red}0

Step 2: Next, we bring {\color{red}30} to the right.

200
3

6 3 0
-\! 6 0 0
\color{red}3 \color{red}0
\color{red}3 \color{red}0

Step 3: Pick a multiple of 3 that is easy to compute but is no larger than 30. Let's pick 3 \times {\color{blue}10} = 30.

We write \color{blue}10 above the box and subtract 3 \times {\color{blue}10} from \color{red}30 inside the box.

200 \color{blue}10
3

6 3 0
-\! 6 0 0
\color{red}3 \color{red}0
\color{red}3 \color{red}0
-\! 3 0
\fbox{0}

We get \fbox{0}, so the division is done.

Step 4: To find the quotient of 630 \div 3, we add the numbers on top of the boxes: 200+10=210

Therefore, 630 \div 3 = 210.

Let's now look at an example with a remainder.

FLAG

100 \fbox{[math]\,\phantom{0}\,[/math]}
4

4 2 1
-\! \fbox{[math]\phantom{0}[/math]} \fbox{[math]\phantom{0}[/math]} \fbox{[math]\phantom{0}[/math]}
\fbox{[math]\phantom{0}[/math]} \fbox{[math]\phantom{0}[/math]}
\fbox{[math]\phantom{0}[/math]} \fbox{[math]\phantom{0}[/math]}
-\!\! \fbox{[math]\phantom{0}[/math]} \fbox{[math]\phantom{0}[/math]}
\fbox{1}

Use the box model above to compute the quotient of 421 \div 4.

EXPLANATION

First, we subtract 4 \times {\color{blue}100} = 400 from 421.

\color{blue}100 \fbox{[math]\,\phantom{0}\,[/math]}
4

4 2 1
-\! \fbox{[math]4[/math]} \fbox{[math]0[/math]} \fbox{[math]0[/math]}
\fbox{[math]\color{red}2[/math]} \fbox{[math]\color{red}1[/math]}
\fbox{[math]\phantom{0}[/math]} \fbox{[math]\phantom{0}[/math]}
-\!\! \fbox{[math]\phantom{0}[/math]} \fbox{[math]\phantom{0}[/math]}
\fbox{1}

Next, we bring \color{red}21 to the right.

\color{blue}100 \fbox{[math]\,\phantom{0}\,[/math]}
4

4 2 1
-\! \fbox{[math]4[/math]} \fbox{[math]0[/math]} \fbox{[math]0[/math]}
\fbox{[math]\color{red}2[/math]} \fbox{[math]\color{red}1[/math]}
\fbox{[math]\color{red}2[/math]} \fbox{[math]\color{red}1[/math]}
-\!\! \fbox{[math]\phantom{0}[/math]} \fbox{[math]\phantom{0}[/math]}
\fbox{1}

Then, we pick a multiple of 4 that is easy to compute but is no larger than {\color{red}21}. Let's pick 4 \times {\color{blue}5} = 20.

We write \color{blue}5 above the box and subtract 4 \times {\color{blue}5} from \color{red}21 inside the box.

100 \color{blue}5
4

4 2 1
-\! \fbox{[math]4[/math]} \fbox{[math]0[/math]} \fbox{[math]0[/math]}
\fbox{[math]\color{red}2[/math]} \fbox{[math]\color{red}1[/math]}
\fbox{[math]\color{red}2[/math]} \fbox{[math]\color{red}1[/math]}
-\!\! \fbox{2} \fbox{0}
\fbox{1}

We can't subtract any further since \fbox{1} is less than 4. Hence, 1 is the remainder.

The quotient is the sum of the numbers on top of the boxes:

100+5=105

Therefore, 421 \div 4 = 105 \,\text{R}\, 1.

FLAG

$100$ $\fbox{$\,\phantom{0}\,$}$
$9$

$9$ $5$ $4$
$-\!$ $\fbox{$\phantom{0}$}$ $\fbox{$\phantom{0}$}$ $\fbox{$\phantom{0}$}$
$\fbox{$\phantom{0}$}$ $\fbox{$\phantom{0}$}$
$\fbox{$\phantom{0}$}$ $\fbox{$\phantom{0}$}$
$-\!\!$ $\fbox{$\phantom{0}$}$ $\fbox{$\phantom{0}$}$
$\fbox{0}$

Use the box model above to compute the value of $954 \div 9.$

a
$104$
b
$105$
c
$110$
d
$108$
e
$106$

$200$ $\fbox{$\,\phantom{0}\,$}$
$4$

$8$ $0$ $9$
$-\!$ $\fbox{$\phantom{0}$}$ $\fbox{$\phantom{0}$}$ $\fbox{$\phantom{0}$}$
$\fbox{$\phantom{0}$}$
$\fbox{$\phantom{0}$}$
$-\!\!$ $\fbox{$\phantom{0}$}$
$\fbox{1}$

Use the box model above to compute the quotient of $809 \div 4.$

a
$200$
b
$204$
c
$203$
d
$202$
e
$201$

20 \fbox{[math]\,\phantom{0}\,[/math]}
6

1 4 4

Laura paid \[math]144 dollars for 6 concert tickets. Using the box model above, find out how much Laura paid for each ticket.

EXPLANATION

To find out how much Laura paid for each ticket, we need to divide 144 by 6.

First, we subtract 6 \times {\color{blue}{20}}=120 from 144.

\color{blue}20
6

1 4 4
-\! 1 2 0
\color{red} \color{red}2 \color{red}4

Next, we bring \color{red}24 to the right.

20
6

1 4 4
-\! 1 2 0
\color{red} \color{red}2 \color{red}4
\!\!\!\!\!{\color{red}2} \!\!\!\!\!\color{red}4

Then, we pick a multiple of 6 that is easy to compute but is no larger than {\color{red}24}. Let's pick 6 \times {\color{blue}4} = 24.

We write \color{blue}4 above the box and subtract 6 \times {\color{blue}4} from \color{red}24 inside the box.

20 \color{blue}4
6

1 4 4
-\! 1 2 0
\color{red}2 \color{red}4
\color{red}2 \color{red}4
-\! 2 4
\fbox{0}

We get \fbox{0}, so the division is done.

The quotient is the sum of the numbers on top of the boxes: 20+4=24

So, 144\div 6 =24

Therefore, Laura paid \[math]24 for each ticket.

FLAG

$100$ $\fbox{$\,\phantom{0}\,$}$
$6$

$6$ $1$ $7$

Use the box model above to find the value of $617 \div 6.$

a
$104\,\textrm{R}\,3$
b
$103\,\textrm{R}\,1$
c
$102\,\textrm{R}\,5$
d
$102\,\textrm{R}\,1$
e
$104\,\textrm{R}\,5$

$50$ $\fbox{$\,\phantom{0}\,$}$
$5$

$2$ $5$ $5$

Henry cuts a $255\,\textrm{in}$ long rope into $5$ pieces of equal length. Using the box model above, find the length of each piece of rope.

a
$53\,\textrm{in}$
b
$55\,\textrm{in}$
c
$57\,\textrm{in}$
d
$51\,\textrm{in}$
e
$65\,\textrm{in}$

Sometimes, we might want to use three (or more) boxes to solve a division problem. This often happens when the dividend in our division problem is large.

Although there is no limit to how many boxes we can use, we'll stick to a maximum of three boxes in this lesson.

Let's see an example.

FLAG

100 20 \fbox{[math]\,\phantom{0}\,[/math]}
6

7 7 8

Use the box model above to find the value of 778 \div 6.

EXPLANATION

First, we subtract 6 \times {\color{blue}100} = 600 from 778.

\color{blue}100
6

7 7 8
-\!\!\! 6 0 0
\color{red}1 \color{red}7 \color{red}8

Next, we bring \color{red}178 to the right and subtract 6 \times {\color{blue}20} = 120 from it.

100 \color{blue}20
6

7 7 8
-\!\!\! 6 0 0
\color{red}1 \color{red}7 \color{red}8
\color{red}1 \color{red}7 \color{red}8
-\!\!\! 1 2 0
\color{red}5 \color{red}8

Next, we bring {\color{red}58} to the right.

100 20
6

7 7 8
-\!\!\! 6 0 0
\color{red}1 \color{red}7 \color{red}8
\color{red}1 \color{red}7 \color{red}8
-\!\!\! 1 2 0
\color{red}5 \color{red}8
\!\!\! \color{red}5 \color{red}8

Finally, we pick a multiple of 6 that is easy to compute but is no larger than {\color{red}58}. Let's pick 6 \times {\color{blue}9} = 54.

We write \color{blue}9 above the box and subtract 6 \times {\color{blue}9} from \color{red}58 inside the box.

100 20 \color{blue}9
6

7 7 8
-\!\!\! 6 0 0
\color{red}1 \color{red}7 \color{red}8
\color{red}1 \color{red}7 \color{red}8
-\!\!\! 1 2 0
\color{red}5 \color{red}8
\!\!\! \color{red}5 \color{red}8
-\!\!\! 5 4
\fbox{4}

We can't subtract any further since \fbox{4} is less than 6. Hence, 4 is the remainder.

The quotient is the sum of the numbers on top of the boxes: 100+20+9=129

Therefore, 778\div 6 = 129 \,\text{R}\, 4.

FLAG

$100$ $20$ $\fbox{$\,\phantom{0}\,$}$
$6$

$7$ $3$ $3$

Use the box model above to find the value of $733 \div 6.$

a
$121 \,\text{R}\, 3$
b
$121$
c
$122 \,\text{R}\, 3$
d
$122 \,\text{R}\, 1$
e
$121 \,\text{R}\, 1$

$100$ $50$ $\fbox{$\,\phantom{0}\,$}$
$4$

$6$ $1$ $2$

Maggy used $612\,\textrm{g}$ of all-purpose flour to make $4$ apple pies. Assuming Maggy used the same weight of flour for each pie, use the box model above to find how much flour she used for each pie.

a
$158\,\textrm{g}$
b
$151\,\textrm{g}$
c
$153\,\textrm{g}$
d
$148\,\textrm{g}$
e
$157\,\textrm{g}$
Flag Content
Did you notice an error, or do you simply believe that something could be improved? Please explain below.
SUBMIT
CANCEL