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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 Question #61 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

646

636

656

6460

Correct answer:

646

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, 38 is the multiplier and 17 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 7

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

Then, we multiply 8 and 1 and add the 5 that was carried

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

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.

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

Next, we multiply 3 and 7

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

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

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

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

$\begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}7\\ \times \phantom{0} 3\space{\,}8 \\ \hline \phantom{\,}1\,3\,6 \\+\, 5 \, 1 \, 0\\ \hline 6\, 4 \, 6\end{array}$

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Example Question #62 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

13110

1311

1301

1321

Correct answer:

1311

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, 23 is the multiplier and 57 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

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

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

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

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

Next, we multiply 2 and 7

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

Then, we multiply 2 and 5and add the 1 that was carried

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

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

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

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Example Question #63 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

6730

67200

6710

6720

Correct answer:

6720

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, 84 is the multiplier and 80 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 0

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

Then, we multiply 4 and 8

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

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

Next, we multiply 8 and 0

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

Then, we multiply 8 and 8

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

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

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

Report an Error

Example Question #1051 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

22440

2244

2234

2254

Correct answer:

2244

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, 44 is the multiplier and 51 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 1

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

Then, we multiply 4 and 5

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

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

Next, we multiply 4 and 1

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

Then, we multiply 4 and 5

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

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

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

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Example Question #65 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

836

8360

846

826

Correct answer:

836

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, 22 is the multiplier and 38 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 8

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

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

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

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

Next, we multiply 2 and 8

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

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

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

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

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

Report an Error

Example Question #66 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

1026

1036

1016

10260

Correct answer:

1026

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, 27 is the multiplier and 38 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 7 and 8

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

Then, we multiply 7 and 3 and add the 5 that was carried

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

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

Next, we multiply 2 and 8

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

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

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

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

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

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Example Question #67 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

1628

1648

1638

16380

Correct answer:

1638

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 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 3 and 6

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

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

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

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

Next, we multiply 6 and 6

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

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

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

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

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

Report an Error

Example Question #91 : How To Multiply

Solve:

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

Possible Answers:

3876

3866

3886

38760

Correct answer:

3876

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, 57 is the multiplier and 68 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 7 and 8

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

Then, we multiply 7 and 6 and add the 5 that was carried

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

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

Next, we multiply 5 and 8

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

Then, we multiply 5 and 6and add the 4 that was carried

$\begin{array}{r}\overset{4 \space{\ }}{\ 6} \space{\,}8\\ \times \phantom{0} 5\space{\,}7 \\ \hline \phantom{\,}4\,7\,6 \\ \, 3 \, 4 \, 0 \, 0\end{array}$

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

$\begin{array}{r}\overset{4 \space{\ }}{\ 6} \space{\,}8\\ \times \phantom{0} 5\space{\,}7 \\ \hline \phantom{\,}4\,7\,6 \\+\, 3 \, 4 \, 0 \, 0\\ \hline 3\, 8 \, 7\, 6\end{array}$

Report an Error

Example Question #92 : How To Multiply

Solve:

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

Possible Answers:

6930

703

693

683

Correct answer:

693

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 21 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 1

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

Then, we multiply 3 and 2

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

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

Next, we multiply 3 and 1

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

Then, we multiply 3 and 2

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

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

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

Report an Error

Example Question #93 : How To Multiply

Solve:

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

Possible Answers:

340

320

3300

330

Correct answer:

330

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 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 3 and 0

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

Then, we multiply 3 and 1

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

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

Next, we multiply 3 and 0

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

Then, we multiply 3 and 1

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

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

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

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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 #61 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #62 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #63 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #1051 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #65 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

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