5th Grade Math - Fluently Multiply Multi-Digit Whole Numbers: CCSS.Math.Content.5.NBT.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{,}5\ \times\phantom{0}8\space{,}5 \ \hline \phantom{,}\end{array}$

Explanation

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 85 is the multiplier and 35 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 5 and 5

$\begin{array}{r}\overset{2 \space{\ }}{\ 3} \space{,}5\ \times \phantom{0} 8\space{,}5\ \hline \phantom{,} ,5\end{array}$

Then, we multiply 5 and 3 and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 3} \space{,}5\ \times \phantom{0} 8\space{,}5 \ \hline \phantom{,}1,7,5\end{array}$

Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{,}5\ \times \phantom{0} 8\space{,}5 \ \hline \phantom{,}1,7,5 \ , 0 \end{array}$

Next, we multiply 8 and 5

$\begin{array}{r}\overset{4 \space{\ }}{\ 3} \space{,}5\ \times \phantom{0} 8\space{,}5 \ \hline \phantom{,}1,7,5 \ , 0 , 0\end{array}$

Then, we multiply 8 and 3and add the 4 that was carried

$\begin{array}{r}\overset{4 \space{\ }}{\ 3} \space{,}5\ \times \phantom{0} 8\space{,}5 \ \hline \phantom{,}1,7,5 \ , 2 , 8 , 0 , 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{4 \space{\ }}{\ 3} \space{,}5\ \times \phantom{0} 8\space{,}5 \ \hline \phantom{,}1,7,5 \+, 2 , 8 , 0 , 0\ \hline 2, 9 , 7, 5\end{array}$

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{,}5\ \times\phantom{0}1\space{,}3 \ \hline \phantom{,}\end{array}$

Explanation

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 13 is the multiplier and 85 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 3 and 5

$\begin{array}{r}\overset{1 \space{\ }}{\ 8} \space{,}5\ \times \phantom{0} 1\space{,}3\ \hline \phantom{,} ,5\end{array}$

Then, we multiply 3 and 8 and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 8} \space{,}5\ \times \phantom{0} 1\space{,}3 \ \hline \phantom{,}2,5,5\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{,}5\ \times \phantom{0} 1\space{,}3 \ \hline \phantom{,}2,5,5 \ , 0 \end{array}$

Next, we multiply 1 and 5

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{,}5\ \times \phantom{0} 1\space{,}3 \ \hline \phantom{,}2,5,5 \ , 5 , 0\end{array}$

Then, we multiply 1 and 8

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{,}5\ \times \phantom{0} 1\space{,}3 \ \hline \phantom{,}2,5,5 \ , 8 , 5 , 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{,}5\ \times \phantom{0} 1\space{,}3 \ \hline \phantom{,}2,5,5 \+, 8 , 5 , 0\ \hline 1, 1 , 0, 5\end{array}$

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{,}2\ \times\phantom{0}7\space{,}8 \ \hline \phantom{,}\end{array}$

Explanation

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 78 is the multiplier and 42 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 8 and 2

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{,}2\ \times \phantom{0} 7\space{,}8\ \hline \phantom{,} ,6\end{array}$

Then, we multiply 8 and 4 and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{,}2\ \times \phantom{0} 7\space{,}8 \ \hline \phantom{,}3,3,6\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{,}2\ \times \phantom{0} 7\space{,}8 \ \hline \phantom{,}3,3,6 \ , 0 \end{array}$

Next, we multiply 7 and 2

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{,}2\ \times \phantom{0} 7\space{,}8 \ \hline \phantom{,}3,3,6 \ , 4 , 0\end{array}$

Then, we multiply 7 and 4and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{,}2\ \times \phantom{0} 7\space{,}8 \ \hline \phantom{,}3,3,6 \ , 2 , 9 , 4 , 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{,}2\ \times \phantom{0} 7\space{,}8 \ \hline \phantom{,}3,3,6 \+, 2 , 9 , 4 , 0\ \hline 3, 2 , 7, 6\end{array}$

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{,}0\ \times\phantom{0}1\space{,}2 \ \hline \phantom{,}\end{array}$

Explanation

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 12 is the multiplier and 10 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 2 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{,}0\ \times \phantom{0} 1\space{,}2 \ \hline \phantom{,} ,0\end{array}$

Then, we multiply 2 and 1

$\begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{,}0\ \times \phantom{0} 1\space{,}2 \ \hline \phantom{,} ,2,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{,}0\ \times \phantom{0} 1\space{,}2 \ \hline \phantom{,} ,2,0 \ , 0 \end{array}$

Next, we multiply 1 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{,}0\ \times \phantom{0} 1\space{,}2 \ \hline \phantom{,} ,2,0 \ , 0 , 0\end{array}$

Then, we multiply 1 and 1

$\begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{,}0\ \times \phantom{0} 1\space{,}2 \ \hline \phantom{,} ,2,0 \ , 1 , 0 , 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{,}0\ \times \phantom{0} 1\space{,}2 \ \hline \phantom{,} ,2,0 \+, 1 , 0 , 0\ \hline 1, 2 , 0\end{array}$

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}1\ \times\phantom{0}4\space{,}0 \ \hline \phantom{,}\end{array}$

Explanation

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 40 is the multiplier and 71 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 0 and 1

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}1\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0\end{array}$

Then, we multiply 0 and 7

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}1\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}1\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0 \ , 0 \end{array}$

Next, we multiply 4 and 1

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}1\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0 \ , 4 , 0\end{array}$

Then, we multiply 4 and 7

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}1\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0 \ , 2, 8, 4 , 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}1\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0 \+, 2, 8, 4 , 0\ \hline 2, 8 , 4, 0\end{array}$

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}8\ \times\phantom{0}4\space{,}0 \ \hline \phantom{,}\end{array}$

Explanation

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 40 is the multiplier and 78 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 0 and 8

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}8\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0\end{array}$

Then, we multiply 0 and 7

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}8\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{,}8\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0 \ , 0 \end{array}$

Next, we multiply 4 and 8

$\begin{array}{r}\overset{3 \space{\ }}{\ 7} \space{,}8\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0 \ , 2 , 0\end{array}$

Then, we multiply 4 and 7and add the 3 that was carried

$\begin{array}{r}\overset{3 \space{\ }}{\ 7} \space{,}8\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0 \ , 3 , 1 , 2 , 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{3 \space{\ }}{\ 7} \space{,}8\ \times \phantom{0} 4\space{,}0 \ \hline \phantom{,} ,0,0 \+, 3 , 1 , 2 , 0\ \hline 3, 1 , 2, 0\end{array}$

Solve:

$\frac{\begin{array}[b]{r}56\ \times\ 17\end{array}}{\ \ \ \ }$

Explanation

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, is the multiplier and is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}^45{\color{Red} 6}\ \times\ {1\color{Red} 7}\end{array}}{\ \ \ \ \ 2}$

Then, we multiply and and add the that was carried

$\frac{\begin{array}[b]{r}^{\color{Red} 4}{\color{Red} 5}6\ \times\ {1\color{Red} 7}\end{array}}{\ \ 392}$

Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a as a place holder in the ones position.

$\frac{\begin{array}[b]{r}5 6\ \times\ 17\end{array}}{\ \ 392 }\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}5 {\color{Red} 6}\ \times\ {\color{Red} 1}7\end{array}}{\ \ 392 }\ {\ \ \ \ \ \ \ \ \ \ 60}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 5} {\color{Black} 6}\ \times\ {\color{Red} 1}7\end{array}}{\ \ 392 }\ {\ \ \ \ \ \ \ \ 560}$

Finally, we add the two products together to find our final answer

$\frac{\begin{array}[b]{r}56\ \times \ \ \ 17\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 392}\ \ +\ {\color{Red} 560} \end{array}}{\ }} \ {\ \ \ \ \ \ \ \ \ \ \ 952}$

$45\times 5=$

$36\times 3=$

$45\times5=225,\ 36\times3=108$

$45\times5=215,\ 36\times3=108$

$45\times5=225,\ 36\times3=208$

$45\times5=220,\ 36\times3=118$

$45\times5=215,\ 36\times3=109$

Explanation

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{,}6\ \times\phantom{0}1\space{,}1 \ \hline \phantom{,}\end{array}$

Explanation

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 11 is the multiplier and 36 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 1 and 6

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{,}6\ \times \phantom{0} 1\space{,}1 \ \hline \phantom{,} ,6\end{array}$

Then, we multiply 1 and 3

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{,}6\ \times \phantom{0} 1\space{,}1 \ \hline \phantom{,} ,3,6\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{,}6\ \times \phantom{0} 1\space{,}1 \ \hline \phantom{,} ,3,6 \ , 0 \end{array}$

Next, we multiply 1 and 6

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{,}6\ \times \phantom{0} 1\space{,}1 \ \hline \phantom{,} ,3,6 \ , 6 , 0\end{array}$

Then, we multiply 1 and 3

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{,}6\ \times \phantom{0} 1\space{,}1 \ \hline \phantom{,} ,3,6 \ , 3 , 6 , 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{,}6\ \times \phantom{0} 1\space{,}1 \ \hline \phantom{,} ,3,6 \+, 3 , 6 , 0\ \hline 3, 9 , 6\end{array}$

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 2} \space{,}6\ \times\phantom{0}3\space{,}6 \ \hline \phantom{,}\end{array}$

Explanation

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 36 is the multiplier and 26 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 6 and 6

$\begin{array}{r}\overset{3 \space{\ }}{\ 2} \space{,}6\ \times \phantom{0} 3\space{,}6\ \hline \phantom{,} ,6\end{array}$

Then, we multiply 6 and 2 and add the 3 that was carried

$\begin{array}{r}\overset{3 \space{\ }}{\ 2} \space{,}6\ \times \phantom{0} 3\space{,}6 \ \hline \phantom{,}1,5,6\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 2} \space{,}6\ \times \phantom{0} 3\space{,}6 \ \hline \phantom{,}1,5,6 \ , 0 \end{array}$

Next, we multiply 3 and 6

$\begin{array}{r}\overset{1 \space{\ }}{\ 2} \space{,}6\ \times \phantom{0} 3\space{,}6 \ \hline \phantom{,}1,5,6 \ , 8 , 0\end{array}$

Then, we multiply 3 and 2and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 2} \space{,}6\ \times \phantom{0} 3\space{,}6 \ \hline \phantom{,}1,5,6 \ , 7 , 8 , 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{1 \space{\ }}{\ 2} \space{,}6\ \times \phantom{0} 3\space{,}6 \ \hline \phantom{,}1,5,6 \+, 7 , 8 , 0\ \hline 9, 3 , 6\end{array}$

Fluently Multiply Multi-Digit Whole Numbers: CCSS.Math.Content.5.NBT.B.5

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