// Numbas version: exam_results_page_options {"name": "Ann's copy of Long division, single digit divisor results in no remainder", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

The simplest case. Divisor is single digit. There is no remainder.

"}, "name": "Ann's copy of Long division, single digit divisor results in no remainder", "variables": {"diff3": {"group": "Ungrouped variables", "name": "diff3", "templateType": "anything", "definition": "dd3-prod3", "description": ""}, "prod1": {"group": "Ungrouped variables", "name": "prod1", "templateType": "anything", "definition": "divisor1*qd1", "description": ""}, "dd2": {"group": "Ungrouped variables", "name": "dd2", "templateType": "anything", "definition": "mod(floor(dividend1/100),10)", "description": ""}, "b0": {"group": "Ungrouped variables", "name": "b0", "templateType": "anything", "definition": "10*diff1+dd0", "description": ""}, "b1": {"group": "Ungrouped variables", "name": "b1", "templateType": "anything", "definition": "10*diff2+dd1", "description": ""}, "qd1": {"group": "Ungrouped variables", "name": "qd1", "templateType": "anything", "definition": "mod(floor(quotient1/10),10)", "description": ""}, "quotient1": {"group": "Ungrouped variables", "name": "quotient1", "templateType": "anything", "definition": "random(ceil(1001/divisor1)..floor(9999/divisor1) except list(100..10000#100))", "description": ""}, "qd0": {"group": "Ungrouped variables", "name": "qd0", "templateType": "anything", "definition": "mod(quotient1,10)", "description": ""}, "dd3": {"group": "Ungrouped variables", "name": "dd3", "templateType": "anything", "definition": "mod(floor(dividend1/1000),10)", "description": ""}, "diff0": {"group": "Ungrouped variables", "name": "diff0", "templateType": "anything", "definition": "b0-prod0", "description": ""}, "prod2": {"group": "Ungrouped variables", "name": "prod2", "templateType": "anything", "definition": "divisor1*qd2", "description": ""}, "qd2": {"group": "Ungrouped variables", "name": "qd2", "templateType": "anything", "definition": "mod(floor(quotient1/100),10)", "description": "

qd2

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

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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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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 (be 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:

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

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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:

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

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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.

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

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

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

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The tens column

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

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

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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:

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

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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.

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

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

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

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The ones column

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

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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:

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

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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.

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

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Now we have run out of digits and we have no remainder. Therefore, $\\var{dividend1}\\div\\var{divisor1}=\\var{quotient1}$.

", "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "unitTests": [], "scripts": {}, "showFeedbackIcon": true, "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "marks": 0, "type": "information"}], "showCorrectAnswer": true, "unitTests": [], "marks": 0, "showFeedbackIcon": true}], "rulesets": {}, "advice": "", "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "

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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