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Common Core: 5th Grade Math : Common Core Math: Grade 5

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

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All Common Core: 5th Grade Math Resources

7 Diagnostic Tests 225 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #22 : 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:

2772

27720

2762

2782

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

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

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.

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

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:

42840

4274

4294

4284

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

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

Report an Error

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:

1176

11760

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

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

Report an Error

Example Question #51 : How To Multiply

Solve:

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

Possible Answers:

890

900

9000

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

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

Report an Error

Example Question #52 : How To Multiply

Solve:

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

Possible Answers:

13200

1320

1310

1330

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

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

Report an Error

Example Question #31 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

3393

34030

3413

3403

Correct answer:

3403

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, 83 is the multiplier and 41 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

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

Then, we multiply 3 and 4

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

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

Next, we multiply 8 and 1

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

Then, we multiply 8 and 4

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

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

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

Report an Error

Example Question #32 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

1485

1475

14850

1495

Correct answer:

1485

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, 55 is the multiplier and 27 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 7

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

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

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

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

Next, we multiply 5 and 7

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

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

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

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

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

Report an Error

Example Question #53 : How To Multiply

Solve:

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

Possible Answers:

4264

4274

42640

4254

Correct answer:

4264

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, 82 is the multiplier and 52 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 2

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

Then, we multiply 2 and 5

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

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

Next, we multiply 8 and 2

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

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

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

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

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

Report an Error

Example Question #54 : How To Multiply

Solve:

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

Possible Answers:

66420

6652

6642

6632

Correct answer:

6642

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, 82 is the multiplier and 81 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 1

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

Then, we multiply 2 and 8

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

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

Next, we multiply 8 and 1

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

Then, we multiply 8 and 8

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

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

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

Report an Error

Example Question #557 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

3696

3706

36960

3686

Correct answer:

3696

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, 56 is the multiplier and 66 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

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

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

$\begin{array}{r}\overset{3 \space{\ }}{\ 6} \space{\,}6\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,}3\,9\,6\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 6} \space{\,}6\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,}3\,9\,6 \\ \, 0 \end{array}$

Next, we multiply 5 and 6

$\begin{array}{r}\overset{3 \space{\ }}{\ 6} \space{\,}6\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,}3\,9\,6 \\ \, 0 \, 0\end{array}$

Then, we multiply 5 and 6and add the 3 that was carried

$\begin{array}{r}\overset{3 \space{\ }}{\ 6} \space{\,}6\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,}3\,9\,6 \\ \, 3 \, 3 \, 0 \, 0\end{array}$

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

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

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All Common Core: 5th Grade Math Resources

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Common Core: 5th Grade Math : Common Core Math: Grade 5

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

All Common Core: 5th Grade Math Resources

Example Questions

Example Question #22 : 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 #51 : How To Multiply

Example Question #52 : How To Multiply

Example Question #31 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #32 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #53 : How To Multiply

Example Question #54 : How To Multiply

Example Question #557 : Number & Operations In Base Ten

All Common Core: 5th Grade Math Resources

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