Fluently Divide Multi-Digit Numbers: CCSS.Math.Content.6.NS.B.2 - Common Core: 6th Grade Math

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Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 93\ 3{\overline{\smash{)}279}}\ -\ 27 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $279\div3=93$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 93}\ {\color{Green} 3}{\overline{\smash{)}279}}\ -\ 27 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$3\times93=279+0$

Simplify.

$3\times93=279$

The correct answer is $3\times 93$

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Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 72\ 2{\overline{\smash{)}144}}\ -\ 14 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $144\div2=72$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 72}\ {\color{Green} 2}{\overline{\smash{)}144}}\ -\ 14 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$2\times72=144+0$

Simplify.

$2\times72=144$

The correct answer is $2\times 72$

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Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 61\ 4{\overline{\smash{)}244}}\ -\ 24 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $244\div4=61$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 61}\ {\color{Green} 4}{\overline{\smash{)}244}}\ -\ 24 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$4\times61=244+0$

Simplify.

$4\times61=244$

The correct answer is $4\times 61$

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Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 11\ 11{\overline{\smash{)}121}}\ -\ 11 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}01{1\ \ }\ -\ \ \ 11\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $121\div11=11$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 11}\ {\color{Green} 11}{\overline{\smash{)}121}}\ -\ 11 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}01{ 1\ \ }\ -\ \ \ 11\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$11\times11=121+0$

Simplify.

$11\times11=121$

The correct answer is $11\times 11$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 54\ 2{\overline{\smash{)}108}}\ -\ 10 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{8\ \ }\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $108\div2=54$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 54}\ {\color{Green} 2}{\overline{\smash{)}108}}\ -\ 10 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 8\ \ }\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$2\times54=108+0$

Simplify.

$2\times54=108$

The correct answer is $2\times 54$

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Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 61\ 2{\overline{\smash{)}122}}\ -\ 12 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{2\ \ }\ -\ \ \ 2\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $122\div2=61$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 61}\ {\color{Green} 2}{\overline{\smash{)}122}}\ -\ 12 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 2\ \ }\ -\ \ \ 2\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$2\times61=122+0$

Simplify.

$2\times61=122$

The correct answer is $2\times 61$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 51\ 3{\overline{\smash{)}153}}\ -\ 15 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{3\ \ }\ -\ \ \ 3\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $153\div3=61$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 51}\ {\color{Green} 3}{\overline{\smash{)}153}}\ -\ 15 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 3\ \ }\ -\ \ \ 3\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$3\times51=153+0$

Simplify.

$3\times51=153$

The correct answer is $3\times 51$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 61\ 3{\overline{\smash{)}183}}\ -\ 18 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{3\ \ }\ -\ \ \ 3\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $183\div3=61$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 61}\ {\color{Green} 3}{\overline{\smash{)}183}}\ -\ 18 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 3\ \ }\ -\ \ \ 3\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$3\times61=183+0$

Simplify.

$3\times61=183$

The correct answer is $3\times 61$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 63\ 3{\overline{\smash{)}189}}\ -\ 18 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $189\div3=63$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 63}\ {\color{Green} 3}{\overline{\smash{)}189}}\ -\ 18 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$3\times63=189+0$

Simplify.

$3\times63=189$

The correct answer is $3\times 63$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 91\ 3{\overline{\smash{)}273}}\ -\ 27 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{3\ \ }\ -\ \ \ 3\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $273\div3=91$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 91}\ {\color{Green} 3}{\overline{\smash{)}273}}\ -\ 27 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 3\ \ }\ -\ \ \ 3\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$3\times91=273+0$

Simplify.

$3\times91=273$

The correct answer is $3\times 91$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 21\ 9{\overline{\smash{)}189}}\ -\ 18 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $189\div9=21$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 21}\ {\color{Green} 9}{\overline{\smash{)}189}}\ -\ 18 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$9\times21=189+0$

Simplify.

$9\times21=189$

The correct answer is $9\times 21$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\ 9{\overline{\smash{)}279}}\ -\ 27 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $279\div9=31$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 31}\ {\color{Green} 9}{\overline{\smash{)}279}}\ -\ 27 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$9\times31=279+0$

Simplify.

$9\times31=279$

The correct answer is $9\times 31$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 21\ 7{\overline{\smash{)}147}}\ -\ 14 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{7\ \ }\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $147\div7=21$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 21}\ {\color{Green} 7}{\overline{\smash{)}147}}\ -\ 14 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 7\ \ }\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$7\times21=147+0$

Simplify.

$7\times21=147$

The correct answer is $7\times 21$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\ 7{\overline{\smash{)}217}}\ -\ 21 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{7\ \ }\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $217\div7=31$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 31}\ {\color{Green} 7}{\overline{\smash{)}217}}\ -\ 21 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 7\ \ }\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$7\times31=217+0$

Simplify.

$7\times31=217$

The correct answer is $7\times 31$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 41\ 7{\overline{\smash{)}287}}\ -\ 28 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{7\ \ }\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $287\div7=41$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 41}\ {\color{Green} 7}{\overline{\smash{)}287}}\ -\ 28 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 7\ \ }\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$7\times41=287+0$

Simplify.

$7\times41=287$

The correct answer is $7\times 41$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 21\ 6{\overline{\smash{)}126}}\ -\ 12 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{6\ \ }\ -\ \ \ 6\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $126\div6=21$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 21}\ {\color{Green} 6}{\overline{\smash{)}126}}\ -\ 12 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 6\ \ }\ -\ \ \ 6\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$6\times21=126+0$

Simplify.

$6\times21=126$

The correct answer is $6\times 21$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\ 6{\overline{\smash{)}186}}\ -\ 18 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{6\ \ }\ -\ \ \ 6\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $186\div6=31$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 31}\ {\color{Green} 6}{\overline{\smash{)}186}}\ -\ 18 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 6\ \ }\ -\ \ \ 6\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$6\times31=186+0$

Simplify.

$6\times31=186$

The correct answer is $6\times 31$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 41\ 6{\overline{\smash{)}246}}\ -\ 24 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{6\ \ }\ -\ \ \ 6\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $246\div6=41$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 41}\ {\color{Green} 6}{\overline{\smash{)}246}}\ -\ 24 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 6\ \ }\ -\ \ \ 6\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$6\times41=246+0$

Simplify.

$6\times41=246$

The correct answer is $6\times 41$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 21\ 8{\overline{\smash{)}168}}\ -\ 16 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{8\ \ }\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $168\div8=21$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 21}\ {\color{Green} 8}{\overline{\smash{)}168}}\ -\ 16 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 8\ \ }\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$8\times21=168+0$

Simplify.

$8\times21=168$

The correct answer is $8\times 21$

Compare your answer with the correct one above

Question

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\ 8{\overline{\smash{)}248}}\ -\ 24 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{8\ \ }\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

Answer

The computation shows that $248\div8=31$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 31}\ {\color{Green} 8}{\overline{\smash{)}248}}\ -\ 24 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 8\ \ }\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$8\times31=248+0$

Simplify.

$8\times31=248$

The correct answer is $8\times 31$

Compare your answer with the correct one above

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