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, 58 is the multiplier and 64 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 4

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

Then, we multiply 8 and 6 and add the 3 that was carried

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

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.

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

Next, we multiply 5 and 4

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

Then, we multiply 5 and 6and add the 2 that was carried

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

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

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

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 35 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 5

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

Then, we multiply 1 and 3

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

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.

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

Next, we multiply 5 and 5

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

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

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

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

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

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, 33 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 3 and 2

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

Then, we multiply 3 and 5

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

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.

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

Next, we multiply 3 and 2

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

Then, we multiply 3 and 5

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

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

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

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, 25 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 5 and 4

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

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

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

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.

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

Next, we multiply 2 and 4

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

Then, we multiply 2 and 8

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

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

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

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, 61 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 1 and 2

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

Then, we multiply 1 and 5

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

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.

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

Next, we multiply 6 and 2

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

Then, we multiply 6 and 5and add the 1 that was carried

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

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

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

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, 66 is the multiplier and 35 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 5

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

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

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

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.

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

Next, we multiply 6 and 5

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

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

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

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

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

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, 57 is the multiplier and 12 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 2

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

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

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

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.

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

Next, we multiply 5 and 2

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

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

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

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

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

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, 34 is the multiplier and 86 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

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

Then, we multiply 4 and 8 and add the 2 that was carried

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

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.

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

Next, we multiply 3 and 6

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

Then, we multiply 3 and 8and add the 1 that was carried

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

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

\(\displaystyle \begin{array}{r}\overset{1 \space{\ }}{\ 8} \space{\,}6\\ \times \phantom{0} 3\space{\,}4 \\ \hline \phantom{\,}3\,4\,4 \\+\, 2 \, 5 \, 8 \, 0\\ \hline 2\, 9 \, 2\, 4\end{array}\)

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, 33 is the multiplier and 67 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

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

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

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

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.

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

Next, we multiply 3 and 7

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

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

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

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

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

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, 72 is the multiplier and 60 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

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

Then, we multiply 2 and 6

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

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.

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

Next, we multiply 7 and 0

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

Then, we multiply 7 and 6

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

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

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