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.

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

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

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

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

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

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.

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

Next, we multiply 5 and 8

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

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

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

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

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

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.

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

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

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

Then, we multiply 3 and 2

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

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.

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

Next, we multiply 3 and 1

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

Then, we multiply 3 and 2

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

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

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

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.

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

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

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

Then, we multiply 3 and 1

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

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.

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

Next, we multiply 3 and 0

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

Then, we multiply 3 and 1

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

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

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

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.

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

For this problem, 62 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

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

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

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

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.

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

Next, we multiply 6 and 8

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

Then, we multiply 6 and 3and add the 4 that was carried

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

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

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

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.

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

For this problem, 81 is the multiplier and 48 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 8

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

Then, we multiply 1 and 4

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

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.

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

Next, we multiply 8 and 8

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

Then, we multiply 8 and 4and add the 6 that was carried

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

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

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

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.

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

For this problem, 32 is the multiplier and 11 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

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

Then, we multiply 2 and 1

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

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.

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

Next, we multiply 3 and 1

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

Then, we multiply 3 and 1

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

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

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

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.

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

For this problem, 11 is the multiplier and 40 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 0

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

Then, we multiply 1 and 4

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

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.

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

Next, we multiply 1 and 0

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

Then, we multiply 1 and 4

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

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

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

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.

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

For this problem, 75 is the multiplier and 88 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 8

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

Then, we multiply 5 and 8 and add the 4 that was carried

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

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.

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

Next, we multiply 7 and 8

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

Then, we multiply 7 and 8and add the 5 that was carried

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

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

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

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.

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

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

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

Then, we multiply 0 and 7

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

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.

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

Next, we multiply 4 and 8

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

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

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

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

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

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.

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

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

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

Then, we multiply 2 and 1

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

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.

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

Next, we multiply 1 and 0

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

Then, we multiply 1 and 1

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

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

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