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, \(\displaystyle 11\) is the multiplier and \(\displaystyle 13\) 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 \(\displaystyle 1\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 3}\\ \times\ {1\color{Red} 1}\end{array}}{\ \ \ \ \ 3}\)

Then, we multiply \(\displaystyle 1\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}3\\ \times\ {1\color{Red} 1}\end{array}}{\ \ \ 13}\)

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 \(\displaystyle 0\) as a place holder in the ones position. 

\(\displaystyle \frac{\begin{array}[b]{r}1 3\\ \times\ 11\end{array}}{\ \ \ 13 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}\)

Next, we multiply \(\displaystyle 1\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}1 {\color{Red} 3}\\ \times\ {\color{Red} 1}1\end{array}}{\ \ \ 13 }\\ {\ \ \ \ \ \ \ \ \ 30}\)

Then, we multiply \(\displaystyle 1\) and \(\displaystyle 1\)

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1} {\color{Black} 3}\\ \times\ {\color{Red} 1}1\end{array}}{\ \ \ 13 }\\ {\ \ \ \ \ \ \ \ 1 30}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}13\\ \times \ \ \ 11\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 13}\\ \ +\ {\color{Red} 130} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 143}\)

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, \(\displaystyle 17\) is the multiplier and \(\displaystyle 21\) 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 \(\displaystyle 7\) and \(\displaystyle 1\)

\(\displaystyle \frac{\begin{array}[b]{r}2{\color{Red} 1}\\ \times\ {1\color{Red} 7}\end{array}}{\ \ \ \ \ 7}\)

Then, we multiply \(\displaystyle 7\) and \(\displaystyle 2\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 2}1\\ \times\ {1\color{Red} 7}\end{array}}{\ \ 147}\)

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 \(\displaystyle 0\) as a place holder in the ones position. 

\(\displaystyle \frac{\begin{array}[b]{r}21\\ \times\ 17\end{array}}{\ \ 147 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}\)

Next, we multiply \(\displaystyle 1\) and \(\displaystyle 1\)

\(\displaystyle \frac{\begin{array}[b]{r}2 {\color{Red} 1}\\ \times\ {\color{Red} 1}7\end{array}}{\ \ 147 }\\ {\ \ \ \ \ \ \ \ \ \ 10}\)

Then, we multiply \(\displaystyle 1\) and \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 2} {\color{Black} 1}\\ \times\ {\color{Red} 1}7\end{array}}{\ \ 147 }\\ {\ \ \ \ \ \ \ \ 210}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}21\\ \times \ \ \ 17\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 147}\\ \ +\ {\color{Red} 210} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 357}\)

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, \(\displaystyle 12\) is the multiplier and \(\displaystyle 36\) 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 \(\displaystyle 2\) and \(\displaystyle 6\)

\(\displaystyle \frac{\begin{array}[b]{r}^13{\color{Red} 6}\\ \times\ {1\color{Red} 2}\end{array}}{\ \ \ \ \ 2}\)

Then, we multiply \(\displaystyle 2\) and \(\displaystyle 3\) and add the \(\displaystyle 1\) that was carried

 \(\displaystyle \frac{\begin{array}[b]{r}^{\color{Red} 1}{\color{Red} 3}6\\ \times\ {1\color{Red} 2}\end{array}}{\ \ \ \ 72}\)

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 \(\displaystyle 0\) as a place holder in the ones position. 

\(\displaystyle \frac{\begin{array}[b]{r}3 6\\ \times\ 12\end{array}}{\ \ \ 72 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}\)

Next, we multiply \(\displaystyle 1\) and \(\displaystyle 6\)

\(\displaystyle \frac{\begin{array}[b]{r}3 {\color{Red} 6}\\ \times\ {\color{Red} 1}2\end{array}}{\ \ \ 72 }\\ {\ \ \ \ \ \ \ \ \ 60}\)

Then, we multiply \(\displaystyle 1\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 3} {\color{Black} 6}\\ \times\ {\color{Red} 1}2\end{array}}{\ \ \ 72 }\\ {\ \ \ \ \ \ \ \ 360}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}36\\ \times \ \ \ 12\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 72}\\ \ +\ {\color{Red} 360} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 432}\)

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, \(\displaystyle 23\) is the multiplier and \(\displaystyle 33\) 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 \(\displaystyle 3\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}3{\color{Red} 3}\\ \times\ {2\color{Red} 3}\end{array}}{\ \ \ \ \ 9}\)

Then, we multiply \(\displaystyle 3\) and \(\displaystyle 3\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 3}3\\ \times\ {2\color{Red} 3}\end{array}}{\ \ \ 99}\)

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 \(\displaystyle 0\) as a place holder in the ones position. 

\(\displaystyle \frac{\begin{array}[b]{r}3 3\\ \times\ 23\end{array}}{\ \ \ 99 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}\)

Next, we multiply \(\displaystyle 2\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}3 {\color{Red} 3}\\ \times\ {\color{Red} 2}3\end{array}}{\ \ \ 99 }\\ {\ \ \ \ \ \ \ \ \ 60}\)

Then, we multiply \(\displaystyle 2\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 3} {\color{Black} 3}\\ \times\ {\color{Red} 2}3\end{array}}{\ \ \ 99 }\\ {\ \ \ \ \ \ \ \ 660}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}33\\ \times \ \ \ 23\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 99}\\ \ +\ {\color{Red} 660} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 759}\)

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, \(\displaystyle 14\) is the multiplier and \(\displaystyle 22\) 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 \(\displaystyle 2\) and \(\displaystyle 4\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 4}\\ \times\ {2\color{Red} 2}\end{array}}{\ \ \ \ \ 8}\)

Then, we multiply \(\displaystyle 2\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}4\\ \times\ {2\color{Red} 2}\end{array}}{\ \ \ 28}\)

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 \(\displaystyle 0\) as a place holder in the ones position. 

\(\displaystyle \frac{\begin{array}[b]{r}1 4\\ \times\ 22\end{array}}{\ \ \ 28 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}\)

Next, we multiply \(\displaystyle 2\) and \(\displaystyle 4\)

\(\displaystyle \frac{\begin{array}[b]{r}1 {\color{Red} 4}\\ \times\ {\color{Red} 2}2\end{array}}{\ \ \ 28 }\\ {\ \ \ \ \ \ \ \ \ 80}\)

Then, we multiply \(\displaystyle 2\) and \(\displaystyle 1\)

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1} {\color{Black} 4}\\ \times\ {\color{Red} 2}2\end{array}}{\ \ \ 28 }\\ {\ \ \ \ \ \ \ \ 280}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}14\\ \times \ \ \ 22\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 28}\\ \ +\ {\color{Red} 280} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 308}\)

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, \(\displaystyle 35\) is the multiplier and \(\displaystyle 43\) 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 \(\displaystyle 5\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}^14{\color{Red} 3}\\ \times\ {3\color{Red} 5}\end{array}}{\ \ \ \ \ 5}\)

Then, we multiply \(\displaystyle 5\) and \(\displaystyle 4\) and add the \(\displaystyle 1\) that was carried

 \(\displaystyle \frac{\begin{array}[b]{r}^{\color{Red} 1}{\color{Red} 4}3\\ \times\ {3\color{Red} 5}\end{array}}{\ \ \ 215}\)

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 \(\displaystyle 0\) as a place holder in the ones position. 

\(\displaystyle \frac{\begin{array}[b]{r}43\\ \times\ 35\end{array}}{\ \ \ 215 }\\ {\ \ \ \ \ \ \ \ \ \ \ \ 0}\)

Next, we multiply \(\displaystyle 3\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}4 {\color{Red} 3}\\ \times\ {\color{Red} 3}5\end{array}}{\ \ \ 215 }\\ {\ \ \ \ \ \ \ \ \ \ 90}\)

Then, we multiply \(\displaystyle 3\) and \(\displaystyle 4\)

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 4} {\color{Black} 3}\\ \times\ {\color{Red} 3}5\end{array}}{\ \ \ 215 }\\ {\ \ \ \ \ \ \ 1290}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}43\\ \times \ \ \ 35\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 215}\\ \ +\ {\color{Red} 1290} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 1505}\)

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, \(\displaystyle 10\) is the multiplier and \(\displaystyle 40\) 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 \(\displaystyle 0\) and \(\displaystyle 0\)

\(\displaystyle \frac{\begin{array}[b]{r}4{\color{Red} 0}\\ \times\ {1\color{Red} 0}\end{array}}{\ \ \ \ \ 0}\)

Then, we multiply \(\displaystyle 0\) and \(\displaystyle 4\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red}4}0\\ \times\ {1\color{Red} 0}\end{array}}{\ \ \ 00}\)

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 \(\displaystyle 0\) as a place holder in the ones position. 

\(\displaystyle \frac{\begin{array}[b]{r}40\\ \times\ 10\end{array}}{\ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}\)

Next, we multiply \(\displaystyle 1\) and \(\displaystyle 0\)

\(\displaystyle \frac{\begin{array}[b]{r}4 {\color{Red} 0}\\ \times\ {\color{Red} 1}0\end{array}}{\ \ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ \ 00}\)

Then, we multiply \(\displaystyle 1\) and \(\displaystyle 4\)

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 4} {\color{Black} 0}\\ \times\ {\color{Red} 1}0\end{array}}{\ \ \ \ 00 }\\ {\ \ \ \ \ \ \ \ 400}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}40\\ \times \ \ \ 10\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 00}\\ \ +\ {\color{Red} 400} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 400}\)

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, \(\displaystyle 43\) is the multiplier and \(\displaystyle 52\) 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 \(\displaystyle 3\) and \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}5{\color{Red} 2}\\ \times\ {4\color{Red} 3}\end{array}}{\ \ \ \ \ 6}\)

Then, we multiply \(\displaystyle 3\) and \(\displaystyle 5\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 5}2\\ \times\ {4\color{Red} 3}\end{array}}{\ \ 156}\)

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 \(\displaystyle 0\) as a place holder in the ones position. 

\(\displaystyle \frac{\begin{array}[b]{r}52\\ \times\ 43\end{array}}{\ \ 156 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}\)

Next, we multiply \(\displaystyle 4\) and \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}5 {\color{Red} 2}\\ \times\ {\color{Red} 4}3\end{array}}{\ \ 156 }\\ {\ \ \ \ \ \ \ \ \ \ 80}\)

Then, we multiply \(\displaystyle 4\) and \(\displaystyle 5\)

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 5} {\color{Black} 2}\\ \times\ {\color{Red} 4}3\end{array}}{\ \ 156 }\\ {\ \ \ \ \ \ \ 2080}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}52\\ \times \ \ \ 43\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 156}\\ \ +\ {\color{Red} 2080} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 2236}\)

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, \(\displaystyle 17\) is the multiplier and \(\displaystyle 56\) 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 \(\displaystyle 7\) and \(\displaystyle 6\)

\(\displaystyle \frac{\begin{array}[b]{r}^45{\color{Red} 6}\\ \times\ {1\color{Red} 7}\end{array}}{\ \ \ \ \ 2}\)

Then, we multiply \(\displaystyle 7\) and \(\displaystyle 5\) and add the \(\displaystyle 4\) that was carried

 \(\displaystyle \frac{\begin{array}[b]{r}^{\color{Red} 4}{\color{Red} 5}6\\ \times\ {1\color{Red} 7}\end{array}}{\ \ 392}\)

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 \(\displaystyle 0\) as a place holder in the ones position. 

\(\displaystyle \frac{\begin{array}[b]{r}5 6\\ \times\ 17\end{array}}{\ \ 392 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}\)

Next, we multiply \(\displaystyle 1\) and \(\displaystyle 6\)

\(\displaystyle \frac{\begin{array}[b]{r}5 {\color{Red} 6}\\ \times\ {\color{Red} 1}7\end{array}}{\ \ 392 }\\ {\ \ \ \ \ \ \ \ \ \ 60}\)

Then, we multiply \(\displaystyle 1\) and \(\displaystyle 5\)

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 5} {\color{Black} 6}\\ \times\ {\color{Red} 1}7\end{array}}{\ \ 392 }\\ {\ \ \ \ \ \ \ \ 560}\)

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

\(\displaystyle \frac{\begin{array}[b]{r}56\\ \times \ \ \ 17\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 392}\\ \ +\ {\color{Red} 560} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 952}\)