// Numbas version: exam_results_page_options {"name": "Division, single digit divisor results in a remainder which is expressed as a fraction - long or short division", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Division, single digit divisor results in a remainder which is expressed as a fraction - long or short division", "tags": [], "metadata": {"description": "

Divisor is single digit. There is a remainder which we express as a fraction. 

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Write the following question down on paper and evaluate it without using a calculator.

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If you are unsure of how to do a question, click on Show steps to see the full working. Then, once you understand how to do the question, click on Try another question like this one to start again.

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this is the quotient without reference to the remainder (so we don't get floating point errors)

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qd2

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$\\var{dividend1}\\div\\var{divisor1}=$[[0]] and [[1]]

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Note, your answer should involve a whole number and a fraction. Put the whole number in the first field and the fraction in the second. For example, if your answer was $4\\frac{3}{5}$ put $4$ in the first field and $3/5$ in the second field.

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Short Division

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We want to calculate $\\var{dividend1}\\div\\var{divisor1}$. By the way, this is the same thing as $\\frac{\\var{dividend1}}{\\var{divisor1}}$. Both of these expressions mean \"how many $\\var{divisor1}$s go into $\\var{dividend1}$?\"

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The short division algorithm allows you to work this out by working from the left to the right of $\\var{dividend1}$ whilst respecting place value. We normally set up the division in the following way:

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$\\var{divisor1} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dividend1}}$

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Note the positions of the numbers!

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We need to know the $\\var{divisor1}$ times tables or write the $\\var{divisor1}$ times tables out (by repeatedly adding $\\var{divisor1}$) so that we can refer to them.

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\\[\\boxed{\\begin{align}1\\times\\var{divisor1}&=\\var{divisor1}\\\\2\\times\\var{divisor1}&=\\var{2*divisor1}\\\\3\\times\\var{divisor1}&=\\var{3*divisor1}\\\\4\\times\\var{divisor1}&=\\var{4*divisor1}\\\\5\\times\\var{divisor1}&=\\var{5*divisor1}\\\\6\\times\\var{divisor1}&=\\var{6*divisor1}\\\\7\\times\\var{divisor1}&=\\var{7*divisor1}\\\\8\\times\\var{divisor1}&=\\var{8*divisor1}\\\\9\\times\\var{divisor1}&=\\var{9*divisor1}\\end{align}}\\]

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The thousands column

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Step 1: The first thing we ask ourselves is, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{dd3}}$?\" (note this $\\var{dd3}$ actually represents $\\var{dd3*1000}$) 

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{dd3}$. So we write $\\color{red}0$ above the $\\var{dd3}$ in the thousands column: Well, $\\var{qd3}\\times \\var{divisor1}=\\var{prod3}$ so $\\var{qd3}$ fit and we write $\\color{red}{\\var{qd3}}$ above the $\\var{dd3}$ in the thousands column:  Well, $\\var{qd3}\\times \\var{divisor1}=\\var{prod3}$ so $\\var{qd3}$ fits and we write $\\color{red}{\\var{qd3}}$ above the $\\var{dd3}$ in the thousands column: 

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$\\begin{array}{r} \\color{red}{\\var{qd3}\\phantom{^\\var{diff3}\\var{qd2}^{\\var{diff2}}\\var{qd1}^{\\var{diff1}}\\var{qd0}}} \\\\[-.5cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\color{green}{\\var{dd3}}^\\phantom{\\var{diff3}}\\var{dd2}^\\phantom{\\var{diff2}}\\var{dd1}^\\phantom{\\var{diff1}}\\var{dd0}} \\end{array}$

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Step 2: We only made it to $\\var{prod3}$ but we were aiming for $\\var{dd3}$, so we were $\\var{diff3}$ away. This is the remainder (what remains to be divided) in the thousands column and we write it in front of the next digit as a superscript in order to represent $\\color{red}{\\var{diff3}}\\var{dd2}$ in the hundreds column. We made it to exactly $\\var{dd3}$, so there is no remainder in the thousands column and you can continue to the next column. However, for the sake of my own sanity (writing this solution), I will put a remainder of $\\color{red}{0}$ in front of the next digit as a superscript in order to represent $\\color{red}{0}\\var{dd2}$ in the hundreds column. You do not need to do this but you can if you wish.

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}\\var{qd2}\\,{^\\var{diff2}}\\var{qd1}\\,{^\\var{diff1}}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^\\color{red}{\\var{diff3}}\\var{dd2}\\,\\phantom{^\\color{red}{\\var{diff2}}}\\var{dd1}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

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The hundreds column

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Step 1: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b2}}$?\" (note this $\\var{b2}$ actually represents $\\var{b2*100}$)

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b2}$. So we write $\\color{red}0$ above the $\\var{dd2}$ in the hundreds column: Well, $\\var{qd2}\\times \\var{divisor1}=\\var{prod2}$ so $\\var{qd2}$ fit and we write $\\color{red}{\\var{qd2}}$ above the $\\var{dd2}$ in the hundreds column:  Well, $\\var{qd2}\\times \\var{divisor1}=\\var{prod2}$ so $\\var{qd2}$ fits and we write $\\color{red}{\\var{qd2}}$ above the $\\var{dd2}$ in the hundreds column: 

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}\\color{red}{\\var{qd2}}\\,\\phantom{^\\var{diff2}\\var{qd1}\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^\\color{green}{\\var{diff3}}\\color{green}{\\var{dd2}}\\,\\phantom{^\\color{red}{\\var{diff2}}}\\var{dd1}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

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Step 2: We only made it to $\\var{prod2}$ but we were aiming for $\\var{b2}$, so we were $\\var{diff2}$ away. This is the remainder (what remains to be divided) in the hundreds column and we write it in front of the next digit as a superscript in order to represent $\\color{red}{\\var{diff2}}\\var{dd1}$ in the tens column. We made it to exactly $\\var{b2}$, so there is no remainder in the hundreds column and you can continue to the next column. However, for the sake of my own sanity (writing this solution), I will put a remainder of $\\color{red}{0}$ in front of the next digit as a superscript in order to represent $\\color{red}{0}\\var{dd1}$ in the tens column. You do not need to do this but you can if you wish.

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,\\phantom{^\\var{diff2}\\var{qd1}\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^{\\var{diff3}}{\\var{dd2}}\\,^\\color{red}{\\var{diff2}}\\var{dd1}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

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The tens column

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Step 1:

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Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b1}}$?\" (note this $\\var{b1}$ actually represents $\\var{b1*10}$)

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b1}$. So we write $\\color{red}0$ above the $\\var{dd1}$ in the tens column: Well, $\\var{qd1}\\times \\var{divisor1}=\\var{prod1}$ so $\\var{qd1}$ fit and we write $\\color{red}{\\var{qd1}}$ above the $\\var{dd1}$ in the tens column:  Well, $\\var{qd1}\\times \\var{divisor1}=\\var{prod1}$ so $\\var{qd1}$ fits and we write $\\color{red}{\\var{qd1}}$ above the $\\var{dd1}$ in the tens column: 

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,\\phantom{^\\var{diff2}}\\color{red}{\\var{qd1}}\\phantom{\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^{\\var{diff3}}{\\var{dd2}}\\,^\\color{green}{\\var{diff2}}\\color{green}{\\var{dd1}}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

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Step 2: We only made it to $\\var{prod1}$ but we were aiming for $\\var{b1}$, so we were $\\var{diff1}$ away. This is the remainder (what remains to be divided) in the tens column and we write it in front of the next digit as a superscript in order to represent $\\color{red}{\\var{diff1}}\\var{dd0}$ in the ones column. We made it to exactly $\\var{b1}$, so there is no remainder in the tens column and you can continue to the next column. However, for the sake of my own sanity (writing this solution), I will put a remainder of $\\color{red}{0}$ in front of the next digit as a superscript in order to represent $\\color{red}{0}\\var{dd0}$ in the ones column. You do not need to do this but you can if you wish.

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,\\phantom{^\\var{diff2}}{\\var{qd1}}\\phantom{\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^{\\var{diff3}}{\\var{dd2}}\\,^{\\var{diff2}}{\\var{dd1}}\\,^\\color{red}{\\var{diff1}}\\var{dd0}}\\end{array}$

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The ones column

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Step 1: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b0}}$?\" 

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b0}$. So we write $\\color{red}0$ above the $\\var{dd0}$ in the ones column: Well, $\\var{qd0}\\times \\var{divisor1}=\\var{prod0}$ so $\\var{qd0}$ fit and we write $\\color{red}{\\var{qd0}}$ above the $\\var{dd0}$ in the ones column:  Well, $\\var{qd0}\\times \\var{divisor1}=\\var{prod0}$ so $\\var{qd0}$ fits and we write $\\color{red}{\\var{qd0}}$ above the $\\var{dd0}$ in the ones column: 

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,\\phantom{^\\var{diff2}}{\\var{qd1}}\\phantom{\\,^\\var{diff1}}\\color{red}{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^{\\var{diff3}}{\\var{dd2}}\\,^{\\var{diff2}}{\\var{dd1}}\\,^\\color{green}{\\var{diff1}}\\color{green}{\\var{dd0}}}\\end{array}$

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Step 2: We only made it to $\\var{prod0}$ but we were aiming for $\\var{b0}$, so we were $\\var{diff0}$ away. This is the remainder (what remains to be divided). 

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,\\phantom{^\\var{diff2}}{\\var{qd1}}\\phantom{\\,^\\var{diff1}}{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^{\\var{diff3}}{\\var{dd2}}\\,^{\\var{diff2}}{\\var{dd1}}\\,^{\\var{diff1}}{\\var{dd0}}}\\end{array} \\begin{array}{r} \\quad \\color{red}{\\text{r}\\,\\var{diff0}}\\\\\\,\\end{array}$

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Now we have run out of digits and we are left with a remainder of $\\var{remainder}$. Note this $\\var{remainder}$ hasn't been divided by $\\var{divisor1}$ yet (that's what remainder means!). Recall that dividing $\\var{remainder}$ by $\\var{divisor1}$ is the same as the fraction $\\frac{\\var{remainder}}{\\var{divisor1}}$ which by removing the common factor of $\\var{gcd}$ from the top and the bottom is equivalent to the fraction $\\frac{\\var{newnum}}{\\var{newden}}$.

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Therefore, $\\var{dividend1}\\div\\var{divisor1}=\\var{quotient1}+\\frac{\\var{newnum}}{\\var{newden}}$, which we usually write without the plus sign as the mixed number $\\var{quotient1}\\frac{\\var{newnum}}{\\var{newden}}$.

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Long Division

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We want to calculate $\\var{dividend1}\\div\\var{divisor1}$. By the way, this is the same thing as $\\frac{\\var{dividend1}}{\\var{divisor1}}$. Both of these expressions mean \"how many $\\var{divisor1}$s go into $\\var{dividend1}$?\"

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The long division algorithm allows you to work this out by working from the left to the right of $\\var{dividend1}$ whilst respecting place value. We normally set up the division in the following way:

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$\\var{divisor1} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dividend1}}$

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Note the positions of the numbers!

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The algorithm (or procedure) seems complicated at first but you might find a mnemonic helps to remember the steps. We work left to right doing the following steps

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    \n
  1. Divide
  2. \n
  3. Multiply
  4. \n
  5. Subtract
  6. \n
  7. Bring down 
  8. \n
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and repeating until we run out of digits. The steps form the acronym DMSB. Popular mnemonics include \"Does McDonalds Sell Burgers?\", \"Dracula Must Suck Blood\" and \"Dead Mice Smell Bad\".

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We need to know the $\\var{divisor1}$ times tables or write the $\\var{divisor1}$ times tables out (by repeatedly adding $\\var{divisor1}$) so that we can refer to them.

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\\[\\boxed{\\begin{align}1\\times\\var{divisor1}&=\\var{divisor1}\\\\2\\times\\var{divisor1}&=\\var{2*divisor1}\\\\3\\times\\var{divisor1}&=\\var{3*divisor1}\\\\4\\times\\var{divisor1}&=\\var{4*divisor1}\\\\5\\times\\var{divisor1}&=\\var{5*divisor1}\\\\6\\times\\var{divisor1}&=\\var{6*divisor1}\\\\7\\times\\var{divisor1}&=\\var{7*divisor1}\\\\8\\times\\var{divisor1}&=\\var{8*divisor1}\\\\9\\times\\var{divisor1}&=\\var{9*divisor1}\\end{align}}\\]

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The thousands column

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D: The first thing we ask ourselves is, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{dd3}}$?\" (note this $\\var{dd3}$ actually represents $\\var{dd3*1000}$) 

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{dd3}$. So we write $\\color{red}0$ above the $\\var{dd3}$ in the thousands column: Well, $\\var{qd3}\\times \\var{divisor1}=\\var{prod3}$ so $\\var{qd3}$ fit and we write $\\color{red}{\\var{qd3}}$ above the $\\var{dd3}$ in the thousands column:  Well, $\\var{qd3}\\times \\var{divisor1}=\\var{prod3}$ so $\\var{qd3}$ fits and we write $\\color{red}{\\var{qd3}}$ above the $\\var{dd3}$ in the thousands column: 

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$\\begin{array}{r} \\color{red}{\\var{qd3}\\phantom{\\var{qd2}\\var{qd1}\\var{qd0}}} \\\\[-.5cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\color{green}{\\var{dd3}}\\var{dd2}\\var{dd1}\\var{dd0}} \\end{array}$

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M: Now since $\\color{green}{\\var{qd3}}\\times \\color{green}{\\var{divisor1}}=\\var{prod3}$ we write $\\color{red}{\\var{prod3}}$ underneath in the thousands column:

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 $\\begin{array}{r} \\color{green}{\\var{qd3}\\phantom{\\var{qd2}\\var{qd1}\\var{qd0}}} \\\\[-.5cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}\\var{dd2}\\var{dd1}\\var{dd0}} \\\\[-.5cm] \\color{red}{\\var{prod3}}\\phantom{555}\\end{array}$

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S: We now do the subtraction, $\\color{green}{\\var{dd3}-\\var{prod3}}$, to determine the remainder (what remains to be divided) in the thousands column. 

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$\\begin{array}{r} {\\var{qd3}\\phantom{\\var{qd2}\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\color{green}{\\var{dd3}}\\var{dd2}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{\\color{green}{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]\\color{red}{\\var{diff3}}\\phantom{555}\\end{array}$

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B: Now we bring the $\\color{green}{\\var{dd2}}$ in the hundreds column down next to the remainder so that it forms $\\var{diff3}\\var{dd2}$.

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$\\begin{array}{r} {\\var{qd3}\\phantom{\\var{qd2}\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}\\color{green}{\\var{dd2}}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}}\\color{red}{\\var{dd2}}\\phantom{55}\\end{array}$

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The hundreds column

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D: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b2}}$?\" (note this $\\var{b2}$ actually represents $\\var{b2*100}$)

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b2}$. So we write $\\color{red}0$ above the $\\var{dd2}$ in the hundreds column: Well, $\\var{qd2}\\times \\var{divisor1}=\\var{prod2}$ so $\\var{qd2}$ fit and we write $\\color{red}{\\var{qd2}}$ above the $\\var{dd2}$ in the hundreds column:  Well, $\\var{qd2}\\times \\var{divisor1}=\\var{prod2}$ so $\\var{qd2}$ fits and we write $\\color{red}{\\var{qd2}}$ above the $\\var{dd2}$ in the hundreds column: 

\n

$\\begin{array}{r} {\\var{qd3}\\color{red}{\\var{qd2}}\\phantom{\\var{qd1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]\\color{green}{\\var{diff3}\\var{dd2}}\\phantom{55}\\end{array}$

\n

M: Now since $\\color{green}{\\var{qd2}}\\times \\color{green}{\\var{divisor1}}=\\var{prod2}$ we write $\\color{red}{\\var{prod2}}$ underneath in the hundreds column:

\n

$\\begin{array}{r} {\\var{qd3}\\color{green}{\\var{qd2}}\\phantom{\\var{qd1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\color{red}{\\var{prod2}}\\phantom{55}\\end{array}$

\n

S: We now do the subtraction, $\\color{green}{\\var{b2}-\\var{prod2}}$, to determine the remainder (what remains to be divided) in the hundreds column. 

\n

$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}\\phantom{\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]\\color{green}{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\underline{\\color{green}{\\var{prod2}}}\\phantom{55}\\\\[-0.5cm]\\color{red}{\\var{diff2}}\\phantom{55}\\end{array}$

\n

B: Now we bring the $\\color{green}{\\var{dd1}}$ in the tens column down next to the remainder so that it forms $\\var{diff2}\\var{dd1}$.

\n

$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}\\phantom{\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}\\color{green}{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{55}\\\\[-0.5cm]\\var{diff2}\\color{red}{\\var{dd1}}\\phantom{5}\\end{array}$

\n

\n

The tens column

\n

D: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b1}}$?\" (note this $\\var{b1}$ actually represents $\\var{b1*10}$)

\n

Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b1}$. So we write $\\color{red}0$ above the $\\var{dd1}$ in the tens column: Well, $\\var{qd1}\\times \\var{divisor1}=\\var{prod1}$ so $\\var{qd1}$ fit and we write $\\color{red}{\\var{qd1}}$ above the $\\var{dd1}$ in the tens column:  Well, $\\var{qd1}\\times \\var{divisor1}=\\var{prod1}$ so $\\var{qd1}$ fits and we write $\\color{red}{\\var{qd1}}$ above the $\\var{dd1}$ in the tens column: 

\n

$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}\\color{red}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{55}\\\\[-0.5cm]\\color{green}{\\var{diff2}\\var{dd1}}\\phantom{5}\\end{array}$

\n

M: Now since $\\color{green}{\\var{qd1}}\\times \\color{green}{\\var{divisor1}}=\\var{prod1}$ we write $\\color{red}{\\var{prod1}}$ underneath in the tens column:

\n

$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}\\color{green}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{55}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\color{red}{\\var{prod1}}\\phantom{5}\\end{array}$

\n

S: We now do the subtraction, $\\color{green}{\\var{b1}-\\var{prod1}}$, to determine the remainder (what remains to be divided) in the tens column. 

\n

$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{55}\\\\[-0.5cm]\\color{green}{\\var{diff2}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{\\color{green}{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm] \\color{red}{\\var{diff1}}\\phantom{5}\\end{array}$

\n

B: Now we bring the $\\color{green}{\\var{dd0}}$ in the ones column down next to the remainder so that it forms $\\var{diff1}\\var{dd0}$.

\n

$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}{\\var{dd1}}\\color{green}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{55}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm] {\\var{diff1}}\\color{red}{\\var{dd0}}\\end{array}$

\n

\n

The ones column

\n

D: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b0}}$?\" 

\n

Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b0}$. So we write $\\color{red}0$ above the $\\var{dd0}$ in the ones column: Well, $\\var{qd0}\\times \\var{divisor1}=\\var{prod0}$ so $\\var{qd0}$ fit and we write $\\color{red}{\\var{qd0}}$ above the $\\var{dd0}$ in the ones column:  Well, $\\var{qd0}\\times \\var{divisor1}=\\var{prod0}$ so $\\var{qd0}$ fits and we write $\\color{red}{\\var{qd0}}$ above the $\\var{dd0}$ in the ones column: 

\n

$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}{\\var{qd1}}\\color{red}{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}{\\var{dd1}}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{55}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm] \\color{green}{\\var{diff1}\\var{dd0}}\\end{array}$

\n

M: Now since $\\color{green}{\\var{qd0}}\\times \\color{green}{\\var{divisor1}}=\\var{prod0}$ we write $\\color{red}{\\var{prod0}}$ underneath in the ones column:

\n

$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}{\\var{qd1}}\\color{green}{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}{\\var{dd1}}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{55}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm]{\\var{diff1}\\var{dd0}}\\\\[-0.5cm]\\color{red}{\\var{prod0}}\\end{array}$

\n

S: We now do the subtraction, $\\color{green}{\\var{b0}-\\var{prod0}}$, to determine the remainder (what remains to be divided) in the ones column. 

\n

$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}{\\var{qd1}}{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}{\\var{dd1}}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{555}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{55}\\\\[-0.5cm]{\\var{diff2}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm]\\color{green}{\\var{diff1}\\var{dd0}}\\\\[-0.5cm]\\underline{\\color{green}{\\var{prod0}}}\\\\[-0.5cm]\\color{red}{\\var{diff0}}\\end{array}$

\n


\n

Now we have run out of digits and we are left with a remainder of $\\var{remainder}$. Note this $\\var{remainder}$ hasn't been divided by $\\var{divisor1}$ yet (that's what remainder means!). Recall that dividing $\\var{remainder}$ by $\\var{divisor1}$ is the same as the fraction $\\frac{\\var{remainder}}{\\var{divisor1}}$ which by removing the common factor of $\\var{gcd}$ from the top and the bottom is equivalent to the fraction $\\frac{\\var{newnum}}{\\var{newden}}$.

\n

\n

Therefore, $\\var{dividend1}\\div\\var{divisor1}=\\var{quotient1}+\\frac{\\var{newnum}}{\\var{newden}}$, which we usually write without the plus sign as the mixed number $\\var{quotient1}\\frac{\\var{newnum}}{\\var{newden}}$.

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