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

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Example Questions

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Example Question #21 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

1074

1064

1054

10640

Correct answer:

1064

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, 14 is the multiplier and 76 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 4 and 6

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

Then, we multiply 4 and 7 and add the 2 that was carried

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

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.

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

Next, we multiply 1 and 6

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

Then, we multiply 1 and 7

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

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

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

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Example Question #22 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

1904

1924

1914

19140

Correct answer:

1914

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, 33 is the multiplier and 58 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 8

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

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

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

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

Next, we multiply 3 and 8

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

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

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

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

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

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Example Question #23 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

5485

5475

5465

54750

Correct answer:

5475

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, 73 is the multiplier and 75 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

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

Then, we multiply 3 and 7 and add the 1 that was carried

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

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

Next, we multiply 7 and 5

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

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

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

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

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

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Example Question #24 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

7480

7490

7470

74800

Correct answer:

7480

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, 88 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 8 and 5

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

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

$\begin{array}{r}\overset{4 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8 \\ \hline \phantom{\,}6\,8\,0\end{array}$

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

Next, we multiply 8 and 5

$\begin{array}{r}\overset{4 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8 \\ \hline \phantom{\,}6\,8\,0 \\ \, 0 \, 0\end{array}$

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

$\begin{array}{r}\overset{4 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8 \\ \hline \phantom{\,}6\,8\,0 \\ \, 6 \, 8 \, 0 \, 0\end{array}$

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

$\begin{array}{r}\overset{4 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 8\space{\,}8 \\ \hline \phantom{\,}6\,8\,0 \\+\, 6 \, 8 \, 0 \, 0\\ \hline 7\, 4 \, 8\, 0\end{array}$

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Example Question #25 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

28710

2871

2881

2861

Correct answer:

2871

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, 33 is the multiplier and 87 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 7

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

Then, we multiply 3 and 8 and add the 2 that was carried

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

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

Next, we multiply 3 and 7

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

Then, we multiply 3 and 8and add the 2 that was carried

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

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

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

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Example Question #26 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

2782

27720

2772

2762

Correct answer:

2772

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, 63 is the multiplier and 44 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 4

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

Then, we multiply 3 and 4 and add the 1 that was carried

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

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

Next, we multiply 6 and 4

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

Then, we multiply 6 and 4and add the 2 that was carried

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

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

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

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Example Question #27 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

4294

4284

42840

4274

Correct answer:

4284

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, 51 is the multiplier and 84 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 4

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

Then, we multiply 1 and 8

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

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

Next, we multiply 5 and 4

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

Then, we multiply 5 and 8and add the 2 that was carried

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

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

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

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Example Question #28 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

11760

1176

1186

1166

Correct answer:

1176

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, 28 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

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

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

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

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

Next, we multiply 2 and 2

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

Then, we multiply 2 and 4

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

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

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

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Example Question #29 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}5\\ \times\phantom{0}2\space{\,}0 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

900

9000

890

910

Correct answer:

900

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, 20 is the multiplier and 45 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 5

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}5\\ \times \phantom{0} 2\space{\,}0 \\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 0 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}5\\ \times \phantom{0} 2\space{\,}0 \\ \hline \phantom{\,} \,0\,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}5\\ \times \phantom{0} 2\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 0 \end{array}$

Next, we multiply 2 and 5

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

Then, we multiply 2 and 4and add the 1 that was carried

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

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

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

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Example Question #30 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

1310

1330

13200

1320

Correct answer:

1320

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 33 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 3

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

Then, we multiply 0 and 3

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

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

Next, we multiply 4 and 3

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

Then, we multiply 4 and 3and add the 1 that was carried

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

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

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

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

Study concepts, example questions & explanations for Common Core: 5th Grade Math

All Common Core: 5th Grade Math Resources

Example Questions

Example Question #21 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #22 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #23 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #24 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #25 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #26 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #27 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #28 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #29 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #30 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

All Common Core: 5th Grade Math Resources

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