To find the quotient, we need to determine how many times $5$ goes into $36.$

We list the multiples of $5$ until we get to one that's larger than $36\mathbin{:}$ \begin{align*} 1 \times 5 & = 5 \\[5pt] 2 \times 5 & = 10 \\[5pt] 3 \times 5 & = 15 \\[5pt] 4 \times 5 & = 20 \\[5pt] 5 \times 5 & = 25 \\[5pt] 6 \times 5 & = 30 \\[5pt] {\color{blue}{7}} \times 5 & = 35 \qquad{\color{green}{\checkmark}}\\[5pt] 8 \times 5 & = 40 \qquad{\color{red}{\times}} \end{align*}

So, $5$ goes into $36$ a total of ${\color{blue}{7}}$ times. Therefore, the quotient is ${\color{blue}{7}}.$

Finally, we can write \[ 36 \div 5 =\,\, \bbox[3pt,Gainsboro]{\color{blue}7} \,\, \textrm{R} \: 1. \]

To find the quotient, we need to determine how many times $3$ goes into $20.$

We list the multiples of $3$ until we get to one that's larger than $20\mathbin{:}$ \begin{align*} 1 \times 3 & = 3 \\[5pt] 2 \times 3 & = 6 \\[5pt] 3 \times 3 & = 9\\[5pt] 4 \times 3 & = 12 \\[5pt] 5 \times 3 & = 15 \\[5pt] {\color{blue}{6}} \times 3 & = 18 \qquad{\color{green}{\checkmark}}\\[5pt] 7 \times 3 & = 21 \qquad{\color{red}{\times}} \end{align*}

So, $3$ goes into $20$ a total of ${\color{blue}{6}}$ times. Therefore, the quotient is ${\color{blue}{6}}.$

Finally, we can write \[ 20 \div 3 =\,\, \bbox[3pt,Gainsboro]{\color{blue}6} \,\, \textrm{R} \: 2. \]

To find the quotient, we need to determine how many times $6$ goes into $21.$

We list the multiples of $6$ until we get to one that's larger than $21\mathbin{:}$ \begin{align*} 1 \times 6 & = 6 \\[5pt] 2 \times 6 & = 12 \\[5pt] {\color{blue}{3}} \times 6 & = 18 \qquad{\color{green}{\checkmark}}\\[5pt] 4 \times 6 & = 24 \qquad{\color{red}{\times}} \end{align*}

So, $6$ goes into $21$ a total of ${\color{blue}{3}}$ times. Therefore, the quotient is ${\color{blue}{3}}.$

Finally, we can write \[ 21 \div 6 =\,\, \bbox[3pt,Gainsboro]{\color{blue}3} \,\, \textrm{R} \: 3. \]

Let's compute the quotient and remainder:

• To find the quotient, we list the multiples of $5$ until we get to one that's larger than $37\mathbin{:}$ \begin{align*} 1\times 5 &= 5 \\[5pt] 2\times 5 &= 10 \\[5pt] &\vdots \\[5pt] {\color{red}{7}}\times 5 &= 35 \qquad{\color{green}{\checkmark}}\\[5pt] 8\times 5 &= 40 \qquad{\color{red}{\times}} \end{align*} Therefore, the quotient is ${\color{red}{7}},$ and we can write the following: \[ 37 \div 5 = {\color{red}{7}}\,\,\textrm R\,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

• To find the remainder, we write the last equation as a multiplication plus the remainder: \[ 37 = 5\times {\color{red}{7}} \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \] This equation simplifies as follows: \[ 37 = 35 \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

Therefore, the remainder must be ${\color{blue}{2}}.$ Finally, we have

\[ 37 \div 5 = {\color{red}7} \, \textrm{R} \, {\color{blue}2}. \]

Let's compute the quotient and remainder:

• To find the quotient, we list the multiples of $4$ until we get to one that's larger than $25{:}$ \begin{align*} 1 \times 4 & = 4 \\[5pt] 2 \times 4 & = 8 \\[5pt] & \vdots \\[5pt] {\color{red}{6}} \times 4 & = 24 \qquad{\color{green}{\checkmark}}\\[5pt] 7 \times 4 & = 28 \qquad{\color{red}{\times}} \end{align*}

Therefore, the quotient is ${\color{red}{6}},$ and we can write the following: \[ 25 \div 4 = {\color{red}{6}}\,\,\textrm R\,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

• To find the remainder, we write the last equation as a multiplication plus the remainder: \[ 25 = 4 \times {\color{red}{6}} \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \] This equation simplifies as follows: \[ 25 = 24 \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

Therefore, the remainder must be ${\color{blue}{1}}.$ Finally, we have \[ 25 \div 4 = \bbox[3pt,Gainsboro]{\color{red}6} \, \textrm{R} \, \bbox[3pt,Gainsboro]{\color{blue}1}. \]

Let's compute the quotient and remainder:

• To find the quotient, we list the multiples of $7$ until we get to one that's larger than $34{:}$ \begin{align*} 1 \times 7 & = 7 \\[5pt] 2 \times 7 & = 14 \\[5pt] 3 \times 7 & = 21 \\[5pt] {\color{red}{4}} \times 7 & = 28 \qquad{\color{green}{\checkmark}}\\[5pt] 5 \times 7 & = 35 \qquad{\color{red}{\times}} \end{align*} Therefore, the quotient is ${\color{red}{4}},$ and we can write the following: \[ 34 \div 7 = {\color{red}{4}}\,\,\textrm R\,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

• To find the remainder, we write the last equation as a multiplication plus the remainder: \[ 34 = 7 \times {\color{red}{4}} \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \] This equation simplifies as follows: \[ 34 = 28 \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

Therefore, the remainder must be ${\color{blue}{6}}.$ Finally, we have \[ 34 \div 7 = {\color{red}4} \, \textrm{R} \, {\color{blue}6}. \]

Let's compute the quotient and remainder:

• To find the quotient, we list the multiples of $5$ until we get to one that's larger than $49{:}$ \begin{align*} 1 \times 5 & = 5 \\[5pt] 2 \times 5 & = 10 \\[5pt] 3 \times 5 & = 15 \\[5pt] 4 \times 5 & = 20 \\[5pt] 5 \times 5 & = 25 \\[5pt] 6 \times 5 & = 30 \\[5pt] 7 \times 5 & = 35 \\[5pt] 8 \times 5 & = 40 \\[5pt] {\color{red}{9}} \times 5 & = 45 \qquad{\color{green}{\checkmark}}\\[5pt] 10 \times 5 & = 50 \qquad{\color{red}{\times}} \end{align*} Therefore, the quotient is ${\color{red}{9}},$ and we can write the following: \[ 49 \div 5 = {\color{red}{9}}\,\,\textrm R\,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

• To find the remainder, we write the last equation as a multiplication plus the remainder: \[ 49 = 5 \times {\color{red}{9}} \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \] This equation simplifies as follows: \[ 49 = 45 \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

Therefore, the remainder must be ${\color{blue}{4}}.$ Finally, we have \[ 49 \div 5 = {\color{red}9} \, \textrm{R} \, {\color{blue}4}. \]

Let's compute the quotient and remainder:

• To find the quotient, we list the multiples of $4$ until we get to one that's larger than $34{:}$ \begin{align*} 1 \times 4 & = 4 \\[5pt] 2 \times 4 & = 8 \\[5pt] 3 \times 4 & = 12 \\[5pt] 4 \times 4 & = 16 \\[5pt] 5 \times 4 & = 20\\[5pt] 6 \times 4 & = 24\\[5pt] 7 \times 4 & = 28\\[5pt] {\color{red}{8}} \times 4 & = 32 \qquad{\color{green}{\checkmark}}\\[5pt] 9 \times 4 & = 36 \qquad{\color{red}{\times}} \end{align*} Therefore, the quotient is ${\color{red}{8}},$ and we can write the following: \[ 34 \div 4 = {\color{red}{8}}\,\,\textrm R\,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

• To find the remainder, we write the last equation as a multiplication plus the remainder: \[ 34 = 4 \times {\color{red}{8}} \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \] This equation simplifies as follows: \[ 34 = 32 \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

Therefore, the remainder must be ${\color{blue}{2}}.$ Finally, we have \[ 34 \div 4 = \bbox[3pt,Gainsboro]{\color{red}8} \, \textrm{R} \, \bbox[3pt,Gainsboro]{\color{blue}2}. \]

Let's compute the quotient and remainder:

• To find the quotient, we list the multiples of $9$ until we get to one that's larger than $47{:}$ \begin{align*} 1 \times 9 & = 9 \\[5pt] 2 \times 9 & = 18 \\[5pt] 3 \times 9 & = 27 \\[5pt] 4 \times 9 & = 36 \\[5pt] {\color{red}{5}} \times 9 & = 45 \qquad{\color{green}{\checkmark}}\\[5pt] 6 \times 9 & = 54 \qquad{\color{red}{\times}} \end{align*}

Therefore, the quotient is ${\color{red}{5}},$ and we can write the following: \[ 47 \div 9 = {\color{red}{5}}\,\,\textrm R\,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

• To find the remainder, we write the last equation as a multiplication plus the remainder: \[ 47 = 9 \times {\color{red}{5}} \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \] This equation simplifies as follows: \[ 47 = 45 \,\,+ \,\,\bbox[3pt, white, border: 1px solid black]{\color{blue}?} \]

Therefore, the remainder must be ${\color{blue}{2}}.$ Finally, we have \[ 47 \div 9 = \bbox[3pt,Gainsboro]{\color{red}5} \, \textrm{R} \, \bbox[3pt,Gainsboro]{\color{blue}2}. \]