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, 45 is the multiplier and 15 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 5

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

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

\(\displaystyle \begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5\end{array}\)

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.

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5 \\ \, 0 \end{array}\)

Next, we multiply 4 and 5

\(\displaystyle \begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5 \\ \, 0 \, 0\end{array}\)

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

\(\displaystyle \begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5 \\ \, 6 \, 0 \, 0\end{array}\)

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

\(\displaystyle \begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5 \\+\, 6 \, 0 \, 0\\ \hline 6\, 7 \, 5\end{array}\)

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, 64 is the multiplier and 56 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

\(\displaystyle \begin{array}{r}\overset{2 \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 6\space{\,}4\\ \hline \phantom{\,} \,4\end{array}\)

Then, we multiply 4 and 5 and add the 2 that was carried

\(\displaystyle \begin{array}{r}\overset{2 \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 6\space{\,}4 \\ \hline \phantom{\,}2\,2\,4\end{array}\)

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.

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 6\space{\,}4 \\ \hline \phantom{\,}2\,2\,4 \\ \, 0 \end{array}\)

Next, we multiply 6 and 6

\(\displaystyle \begin{array}{r}\overset{3 \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 6\space{\,}4 \\ \hline \phantom{\,}2\,2\,4 \\ \, 6 \, 0\end{array}\)

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

\(\displaystyle \begin{array}{r}\overset{3 \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 6\space{\,}4 \\ \hline \phantom{\,}2\,2\,4 \\ \, 3 \, 3 \, 6 \, 0\end{array}\)

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

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

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, 30 is the multiplier and 43 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

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

Then, we multiply 0 and 4

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

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.

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

Next, we multiply 3 and 3

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}3\\ \times \phantom{0} 3\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 9 \, 0\end{array}\)

Then, we multiply 3 and 4

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}3\\ \times \phantom{0} 3\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 1\, 2\, 9 \, 0\end{array}\)

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

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}3\\ \times \phantom{0} 3\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\+\, 1\, 2\, 9 \, 0\\ \hline 1\, 2 \, 9\, 0\end{array}\)

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, 16 is the multiplier and 73 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 3

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

Then, we multiply 6 and 7 and add the 1 that was carried

\(\displaystyle \begin{array}{r}\overset{1 \space{\ }}{\ 7} \space{\,}3\\ \times \phantom{0} 1\space{\,}6 \\ \hline \phantom{\,}4\,3\,8\end{array}\)

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.

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

Next, we multiply 1 and 3

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

Then, we multiply 1 and 7

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

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

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

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, 40 is the multiplier and 71 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 1

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}1\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\end{array}\)

Then, we multiply 0 and 7

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}1\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0\end{array}\)

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.

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}1\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 0 \end{array}\)

Next, we multiply 4 and 1

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}1\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 4 \, 0\end{array}\)

Then, we multiply 4 and 7

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

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

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

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, 42 is the multiplier and 74 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 4

\(\displaystyle \begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}4\\ \times \phantom{0} 4\space{\,}2 \\ \hline \phantom{\,} \,8\end{array}\)

Then, we multiply 2 and 7

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

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.

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

Next, we multiply 4 and 4

\(\displaystyle \begin{array}{r}\overset{1 \space{\ }}{\ 7} \space{\,}4\\ \times \phantom{0} 4\space{\,}2 \\ \hline \phantom{\,}1\,4\,8 \\ \, 6 \, 0\end{array}\)

Then, we multiply 4 and 7and add the 1 that was carried

\(\displaystyle \begin{array}{r}\overset{1 \space{\ }}{\ 7} \space{\,}4\\ \times \phantom{0} 4\space{\,}2 \\ \hline \phantom{\,}1\,4\,8 \\ \, 2 \, 9 \, 6 \, 0\end{array}\)

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

\(\displaystyle \begin{array}{r}\overset{1 \space{\ }}{\ 7} \space{\,}4\\ \times \phantom{0} 4\space{\,}2 \\ \hline \phantom{\,}1\,4\,8 \\+\, 2 \, 9 \, 6 \, 0\\ \hline 3\, 1 \, 0\, 8\end{array}\)

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, 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

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

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

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

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.

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

Next, we multiply 3 and 7

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

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

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

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

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

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, 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

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

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

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

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.

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

Next, we multiply 2 and 7

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

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

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

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

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

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, 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

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

Then, we multiply 4 and 8

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

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.

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

Next, we multiply 8 and 0

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

Then, we multiply 8 and 8

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

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

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

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, 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

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

Then, we multiply 4 and 5

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

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.

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

Next, we multiply 4 and 1

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

Then, we multiply 4 and 5

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

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

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