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

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 51\ 9{\overline{\smash{)}459}}\ -\ 45 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

$9\times 51$

$9\times 45$

$9\times 50$

$9\times 40$

$9\times 60$

Explanation

The computation shows that $459\div9=51$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 51}\ {\color{Green} 9}{\overline{\smash{)}459}}\ -\ 45 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 9\ \ }\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$9\times51=459+0$

Simplify.

$9\times51=459$

The correct answer is $9\times 51$

Use the computation shown to find the products:

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

$3\times 93$

$3\times 3$

$3\times 90$

$3\times 279$

$3\times 99$

Explanation

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 $3\times 93$

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 81\ 4{\overline{\smash{)}324}}\ -\ 32 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

$4\times 81$

$4\times 91$

$4\times 101$

$4\times 32$

$4\times 82$

Explanation

The computation shows that $324\div4=81$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 81}\ {\color{Green} 4}{\overline{\smash{)}324}}\ -\ 32 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$4\times81=324+0$

Simplify.

$4\times81=324$

The correct answer is $4\times 81$

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 91\ 4{\overline{\smash{)}364}}\ -\ 36 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

$4\times 91$

$4\times 101$

$4\times 36$

$4\times 63$

$4\times 81$

Explanation

The computation shows that $364\div4=91$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 91}\ {\color{Green} 4}{\overline{\smash{)}364}}\ -\ 36 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$4\times91=364+0$

Simplify.

$4\times91=364$

The correct answer is $4\times 91$

Use the computation shown to find the products:

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

$7\times 41$

$7\times 21$

$7\times 31$

$7\times 28$

$7\times 32$

Explanation

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 $7\times 41$

Use the computation shown to find the products:

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

$6\times 41$

$6\times 32$

$6\times 36$

$6\times 21$

$6\times 31$

Explanation

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 $6\times 41$

Use the computation shown to find the products:

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

$2\times 61$

$2\times 63$

$2\times 68$

$3\times 36$

$2\times 76$

Explanation

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 $2\times 61$

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 41\ 8{\overline{\smash{)}328}}\ -\ 32 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{8\ \ }\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

$8\times 41$

$8\times 27$

$8\times 51$

$8\times 31$

$8\times 38$

Explanation

The computation shows that $328\div8=41$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 41}\ {\color{Green} 8}{\overline{\smash{)}328}}\ -\ 32 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 8\ \ }\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$8\times41=328+0$

Simplify.

$8\times41=328$

The correct answer is $8\times 41$

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 51\ 4{\overline{\smash{)}204}}\ -\ 20 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

$4\times 51$

$4\times 61$

$4\times 41$

$4\times 20$

$4\times 24$

Explanation

The computation shows that $204\div4=51$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 51}\ {\color{Green} 4}{\overline{\smash{)}204}}\ -\ 20 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 4\ \ }\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$4\times51=204+0$

Simplify.

$4\times51=204$

The correct answer is $4\times 51$

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 61\ 5{\overline{\smash{)}305}}\ -\ 30 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

$5\times 61$

$5\times 81$

$5\times 31$

$5\times 30$

$5\times 63$

Explanation

The computation shows that $305\div5=61$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 61}\ {\color{Green} 5}{\overline{\smash{)}305}}\ -\ 30 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$5\times61=305+0$

Simplify.

$5\times61=305$

The correct answer is $5\times 61$

Fluently Divide Multi-Digit Numbers: CCSS.Math.Content.6.NS.B.2

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