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, 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

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

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

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

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.

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

Next, we multiply 8 and 5

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

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

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

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

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

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 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

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

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

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

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.

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

Next, we multiply 3 and 7

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

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

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

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

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

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, 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

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

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

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

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.

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

Next, we multiply 6 and 4

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

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

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

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

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

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, 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

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

Then, we multiply 1 and 8

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

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.

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

Next, we multiply 5 and 4

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

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

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

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

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

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, 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

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

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

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

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.

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

Next, we multiply 2 and 2

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

Then, we multiply 2 and 4

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

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

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

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, 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

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

Then, we multiply 0 and 4

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

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.

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

Next, we multiply 2 and 5

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

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

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

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

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

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 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

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

Then, we multiply 0 and 3

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

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.

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

Next, we multiply 4 and 3

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

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

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

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

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

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, 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

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

Then, we multiply 3 and 4

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

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.

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

Next, we multiply 8 and 1

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

Then, we multiply 8 and 4

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

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

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

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, 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

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

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

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

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.

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

Next, we multiply 5 and 7

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

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

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

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

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

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, 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

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

Then, we multiply 2 and 5

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

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.

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

Next, we multiply 8 and 2

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

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

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

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

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