Let's consider the following division problem model:

This model represents the division problem

22 \div 5 = {\color{red}{4}} \, \textrm{R} \: {\color{blue}{2}}\,.

We can express this division as a multiplication problem. However, as well as multiplying, we must add the remainder:

22 = 5 \times {\color{red}{4}} + {\color{blue}{2}}

We can also swap the order of the multiplication as follows:

22 = {\color{red}{4}} \times 5 + {\color{blue}{2}}

FLAG

Consider the following division problem: 14 \div 4 = 3 \, \textrm{R} \: 2

Which of the following problems are equivalent to our division problem?

  1. 14 = 4\times 3 + 2
  2. 14 = 4\times 2 + 3
  3. 14 = 3\times 4 + 2
EXPLANATION

First, we express our division as a multiplication problem. However, as well as multiplying, we add the remainder.

14 = 4\times 3 + {\color{blue}{2}}

We can swap the order of the multiplication, as follows:

14 = 3\times 4 + {\color{blue}{2}}

Therefore, the correct answer is "I and III only."

FLAG

Consider the following division problem: \[ 33 \div 4 = 8 \, \textrm{R} \: 1 \]

Which of the following problems are equivalent to our division problem?

  1. $33 = 4\times 1 + 8$
  2. $33 = 8\times 4 + 1$
  3. $33 = 4\times 8 + 1$
a
II and III only
b
II only
c
I and III only
d
I only
e
III only

Consider the following division problem: \[ 26 \div 7 = 3 \, \textrm{R} \: 5 \]

Which of the following problems are equivalent to our division problem?

  1. $26 = 7\times 3 + 5$
  2. $26 = 3\times 7 + 5$
  3. $26 = 7\times 5 + 3$
a
I only
b
III only
c
II only
d
I and III only
e
I and II only

How would we find the remainder in the division problem below?

27 \div 6 = 4 \, \textrm{R} \: \fbox{[math]\phantom{A}[/math]}

First, we express our division as a multiplication problem. However, as well as multiplying, we add the remainder.

27 = 6 \times 4 \,\,+\,\, \bbox[3pt, white, border: 1px solid black]{\color{blue}?}

Now, since 6 \times 4 = 24, the above equation becomes

27 = 24 \,\, +\,\, \bbox[3pt, white, border: 1px solid black]{\color{blue}?}.

So, the remainder must be {\color{blue}{3}}.

27 = 24 \,+\, \bbox[3pt, white, border: 1px solid black]{\color{blue}3}

Finally, we can write

27 \div 6 = 4 \, \textrm{R} \, \bbox[3pt, white, border: 1px solid black]{\color{blue}3}.

FLAG

Consider the following division problem with an unknown remainder. 23 \div 5 = 4 \, \textrm{R} \: \bbox[3pt, white, border: 1px solid black]{\color{blue}?}

We can rewrite this division problem as follows:

23 = 5\times 4 \,\,+\,\, \bbox[3pt, white, border: 1px solid black]{\color{blue}?}

What is the unknown remainder?

EXPLANATION

Since 5\times 4 = 20, our equation becomes

23 = 20 \,\,+\,\, \bbox[3pt, white, border: 1px solid black]{\color{blue}?}.

So, the remainder must be {\color{blue}{3}}.

23 = 20 \,\,+\,\, \bbox[3pt, white, border: 1px solid black]{\color{blue}3}

Finally, we can write

23 \div 5 = 4 \, \textrm{R} \, \bbox[3pt, white, border: 1px solid black]{\color{blue}3}.

FLAG

Consider the following division problem with an unknown remainder. \[ 29 \div 10 = 2 \, \textrm{R} \: \bbox[3pt, white, border: 1px solid black]{\color{blue}?} \] We can rewrite this division problem as follows:

a
b
c
d
e

Consider the following division problem with an unknown remainder. \[ 29 \div 4 = 7 \, \textrm{R} \: \bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

We can rewrite this division problem as follows:

\[ 29 = 4\times 7 \,\,+\,\, \bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

What is the unknown remainder?

a
$5$
b
$0$
c
$2$
d
$3$
e
$1$

Consider the following division problem with an unknown remainder. \[ 70\div 9 = 7 \, \textrm{R} \: \bbox[3pt, white, border: 1px solid black]{\color{blue}?} \] We can rewrite this division problem as follows:

a
b
c
d
e

Find the remainder in the division problem below. 35 \div 6 = 5 \, \textrm{R} \: \fbox{[math]\phantom{A}[/math]}

EXPLANATION

First, we express our division as a multiplication problem. However, as well as multiplying, we add the remainder.

35 = 6\times 5 \,\,+\,\, \bbox[3pt, white, border: 1px solid black]{\color{blue}?}

Now, since 6\times 5 = 30, the above equation becomes

35 = 30 \,\,+\,\, \bbox[3pt, white, border: 1px solid black]{\color{blue}?}.

So, the remainder must be {\color{blue}{5}}.

35 = 30 \,\,+\,\, \bbox[3pt, white, border: 1px solid black]{\color{blue}5}

Finally, we can write

35 \div 6 = 5 \, \textrm{R} \, \bbox[3pt, white, border: 1px solid black]{\color{blue}5}.

FLAG

Find the remainder in the division problem below.

a
b
c
d
e

Find the remainder in the division problem below.

a
b
c
d
e

Find the remainder in the division problem below. \[ 91 \div 8 = 11 \, \textrm{R} \: \fbox{$\phantom{A}$} \]

a
$3$
b
$11$
c
$2$
d
$1$
e
$8$
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