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 

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