Vertical Multiplication: 3-digit by 2-digit multiplication

To solve this problem, we'll follow these steps:

• Step 1: Identify the operation given.
• Step 2: Calculate $117 \times 10$.
• Step 3: Find the corresponding choice from the answers provided.

Now, let's work through each step:
Step 1: The operation involves calculating $117 \times 10$. When multiplying by 10, you retain the original number and append a zero to the end because multiplication by 10 effectively shifts one place value to the left.
Step 2: Calculating $117 \times 10$, we find:
$117 \times 10 = 1170$
This result can easily be verified by observing its pattern and understanding the arithmetic rule when multiplying by 10.
Step 3: The question provides multiple choices and our answer $1170$ matches with choice 2.

Therefore, the solution to the problem is $1170$.

To solve this problem, we need to calculate the product of $130$ and $16$ using the long multiplication method:

• Write down the numbers in vertical form:
      $130 \\times 16$
• Multiply $130$ by the unit digit ($6$) of $16$:
      $130 \times 6 = 780$
• Multiply $130$ by the ten digit ($1$, but representing ten) of $16$:
      $130 \times 10 = 1300$
• Add the two partial products together, aligning them according to place value:
         780
    + 1300
    ---------
• Sum the numbers:
      $2080$

Thus, the product of $130$ and $16$ is $\textbf{2080}$.

Review the provided choices and see that the correct answer is $\boxed{2080}$.

Therefore, the solution to the problem is $\textbf{2080}$.

To solve the problem of multiplying 191 by 10, we will follow these steps:

• Step 1: Recognize that multiplying a number by 10 is equivalent to shifting all digits of the number one place value to the left and adding a zero to the end.
• Step 2: Apply this rule to the number 191.
• Step 3: Perform the multiplication: $191 \times 10 = 1910$.

Let's execute the steps:

When we multiply 191 by 10, each digit in 191 is shifted one place to the left, and we append a zero to the end, resulting in $1910$.

Therefore, the solution to the problem is $1910$.

To solve this problem, we'll follow these steps:

• Step 1: Calculate the product of $122$ and the units digit of $17$.
• Step 2: Calculate the product of $122$ and the tens digit of $17$, considering the place value.
• Step 3: Sum the intermediate products to obtain the final product.

Now, let's work through each step:
Step 1: Multiply $122$ by $7$.
$\begin{array}{c} 122 \\ \times\ 7 \\ \hline 854 \\ \end{array}$
Step 2: Multiply $122$ by $10$ (which is equivalent to multiplying by $1$ and then shifting the result one position to the left).
$\begin{array}{c} 122 \\ \times\ 10 \\ \hline 1220 \\ \end{array}$
Step 3: Add the partial products together.
$\begin{array}{c} 854 \\ +1220 \\ \hline 2074 \\ \end{array}$

Therefore, the solution to the problem is $2074$.

To solve this multiplication problem, we will perform vertical multiplication:

• First, multiply the units digit of $30$ by $151$:
  $\phantom{000}151 \times 0 = 0$, which results in a row of zero due to multiplication by zero.
• Next, multiply the tens digit of $30$ (which is $3$) by each digit in $151$:
  $\phantom{000}151 \times 3 = 453$. When considering place value, this becomes $4530$ because we are actually multiplying by $30$ which is $3 \times 10$.
• Add the two results together:
  $\phantom{000}000$ (result from $151 \times 0$)
  $+ 4530$ (result from $151 \times 3$)
  $\underline{4530}$

The product of $151$ multiplied by $30$ is $\boxed{4530}$.

Therefore, the solution to the problem is $\boxed{4530}$ which matches with the given correct answer choice.

To solve this problem, we'll use vertical multiplication:

• Step 1: Write the multiplication as follows:

$\begin{array}{c} \phantom{+}131 \\ \times \phantom{1}61 \\ \hline \end{array}$

• Step 2: Multiply 131 by the units digit of 61 (which is 1):

$\begin{array}{c} \phantom{+}131 \\ \times \phantom{1}61 \\ \hline \phantom{00}131 \\ \end{array}$

• Step 3: Multiply 131 by the tens digit of 61 (which is 6), and shift one position to the left:

$\begin{array}{c} \phantom{+}131 \\ \times \phantom{1}61 \\ \hline \phantom{0}7860 \\ + \phantom{00}131 \\ \hline \end{array}$

• Step 4: Add the two results together:

$\begin{array}{c} \phantom{+}131 \\ \times \phantom{1}61 \\ \hline \phantom{0}7860 \\ + \phantom{00}131 \\ \hline \phantom{00}7991 \\ \end{array}$

Therefore, the product of 131 and 61 is $7991$.

This matches choice 3 in the given options.

To solve this problem, we will use vertical multiplication. Let's break this down into clear steps:

• Step 1: Multiply the units digit of 31 (which is 1) by the entire number 120.
• Step 2: Multiply the tens digit of 31 (which is 3) by the entire number 120, and add a zero to the end of this result to account for the position of the digit.
• Step 3: Add both products to obtain the final result.

Step 1: Multiply 120 by 1:

$120 \times 1 = 120$

Step 2: Multiply 120 by 3:

$120 \times 3 = 360$

Since this multiplication is coming from the tens place, we add a zero, resulting in 3600.

Step 3: Add the results from Step 1 and Step 2:

$120 + 3600 = 3720$

Therefore, the product of $120$ and $31$ is $\textbf{3720}$.

To solve this problem, follow these steps:

• Step 1: Multiply 111 and 70 using vertical multiplication.

• Step 2: Confirm the calculation by reviewing each step of the multiplication process.

Let's begin:

Step 1: Vertical Multiplication
We'll perform the multiplication of 111 by 70 using vertical multiplication:

$\quad\quad\quad\quad 111$
$\times\quad\quad\quad\; ~~70$
$\quad\quad\quad\;\_\_\_\_\_$
$\quad\quad\quad~~7770$

Explanation:

- We multiply 111 by 0 from 70, giving 0.
- We then multiply 111 by 7 (tens digit of 70). Add a zero to the result of this step to account for its place value:

$\quad\quad 111 \times 7 = 777 \\ \overbrace{\phantom{0}}^{\text{Place value adjustment}} \\ \quad\quad \Rightarrow 777 \times 10 = 7770$

Step 2: Verify the Calculation

By verifying each multiplication step and ensuring correct place value management, the calculation confirms that the product is indeed 7770.

Therefore, the solution to the problem is $x = 7770$.

To solve this multiplication problem, 133 multiplied by 19, using long multiplication, follow these steps:

• Step 1: Write the numbers 133 and 19, aligning them vertically with the larger number on top.
          133
  ×        19
• Step 2: Multiply 133 by the unit digit of 19 (which is 9).
           133
  ×          9
  –––––––––––
         1197 (this is 9 times 133)
• Step 3: Multiply 133 by the tens digit of 19 (which is 1). Remember that because it is the tens digit, we need to add a zero to the end of this product.
           133
  ×        10
  –––––––––––
        1330 (this is 10 times 133)
• Step 4: Add the two results from steps 2 and 3 together.
           1197 (result from step 2)
  +       1330 (result from step 3)
  –––––––––––
         2527

Therefore, the product of 133 and 19 is $2527$.

To solve this problem, we'll find the product of 191 and 20 using vertical multiplication.

Here are the detailed steps:

• Step 1: Multiply 191 by 0 (the units digit of 20).
  Since any number multiplied by 0 is 0, this step results in:
  $191 \times 0 = 0$
• Step 2: Multiply 191 by 2 (the tens digit of 20).
  Let's set up the multiplication:
  $\begin{array}{c} & 1 9 1 \\ \times & 2 \\ \hline & 3 8 2 \\ \end{array}$
• Step 3: Account for the place value of the 2 in the tens position by adding a zero to the end of the product:
  The result is $3820$.
• Step 4: Add the partial products:
  Since the first partial product was 0, we add:
  $0 + 3820 = 3820$

Therefore, the solution to the problem is $3820$, which corresponds to choice 3 in the provided list.

To solve the problem of multiplying $131 \times 11$, follow these steps:

• First, multiply 131 by the units digit of 11 (1):
• $131 \times 1 = 131$
• Next, multiply 131 by the tens digit of 11 (1), but remember to shift by one place to the left as it's a tens position:
• $131 \times 10 = 1310$
• Finally, sum the two products (considering the place values correctly):
• $131 + 1310 = 1441$

Thus, the solution to the multiplication $131 \times 11$ is $\mathbf{1441}$.

To solve the multiplication of $126$ and $17$, we will use vertical multiplication:

• Write $126$ above $17$, aligning them properly according to decimal place values.
• Multiply each digit of the top number by each digit of the bottom number, starting from the lowest place value (ones).
• First, multiply $126$ by $7$ (the ones digit of $17$): $126 \times 7:$

  126
  ×  7
  -----
   882
  

  This yields $882$.
• Next, multiply $126$ by $1$ (tens digit of $17$) with proper adjustment (shift one position to the left): $126 \times 10:$

  126
  ×10
  -----
  1260
  

  This results in $1260$.
• Sum both intermediate products: $882 + 1260 = 2142$

Therefore, the final product of $126 \times 17$ is $2142$.

To solve the problem, we need to multiply 142 by 16 using vertical multiplication:

• Start by multiplying the units digit of 16 (which is 6) by 142:
  $142 \times 6 = 852$

• Then, multiply the tens digit of 16 (which is 1, but placed in the tens position, effectively being 10) by 142:
  $142 \times 10 = 1420$

• Now, add these two results together:
  $\begin{aligned} & 852 \\ + & 1420 \\ \hline & 2272 \end{aligned}$

Thus, the result of multiplying 142 by 16 is $2272$.

Therefore, the solution to the problem is $2272$.

To solve the multiplication problem for $x$, we will calculate it directly through division.

Our problem states:

We can isolate $x$ by dividing both sides of the equation by $116$:

Calculating the division, we find:

The correct solution is therefore $x = 10$. This corresponds to choice 1, which is the answer given as $1160$.

To solve this problem, we will use the vertical multiplication method for multiplying the numbers 102 and 25.

• Step 1: Align 102 and 25 vertically, ensuring 102 is above 25 because it has more digits.
• Step 2: Start by multiplying the digit 5 (the ones place of 25) by each digit of 102 from right to left.

First, multiply 5 by 2 (the units digit of 102):
$5 \times 2 = 10$. Write down 0 and carry over 1.

Next, multiply 5 by 0 (the tens digit of 102) and add the carry-over:
$5 \times 0 + 1 = 1$. Write down 1.

Finally, multiply 5 by 1 (the hundreds digit of 102):
$5 \times 1 = 5$. The result is 510.

• Step 3: Now move to the digit 2 (the tens place of 25) and repeat the process.

Because this is the tens place, we write a zero beneath the 0 from our previous calculation (indenting by one place to the left).

Multiply 2 by 2:
$2 \times 2 = 4$. Write down 4.

Next, multiply 2 by 0:
$2 \times 0 = 0$. Write down 0.

Finally, multiply 2 by 1:
$2 \times 1 = 2$. The result is 2040. Place this beneath the 510, indented correctly.

• Step 4: Add these two results together.

$\begin{array}{c} 510 \\ +2040 \\ \hline 2550 \\ \end{array}$

The sum is 2550. Thus, the product of 102 and 25 is $2550$.

Therefore, the solution to the problem is $2550$.

Let's solve this multiplication problem step by step:

• Step 1: Multiply 231 by the units digit of 21, which is 1. $231 \times 1 = 231$
• Step 2: Multiply 231 by the tens digit of 21, which is 2 (representing 20) $231 \times 2 = 462$ Since it's in the tens place, we append a zero, resulting in 4620.
• Step 3: Add the results of Step 1 and Step 2 together: $231 + 4620 = 4851$

Therefore, the solution to the problem is $4851$.

To solve this problem, we'll multiply the two numbers 341 and 22 using the vertical multiplication technique.

• Step 1: Arrange the numbers vertically, with 341 on top and 22 below it.
• Step 2: Multiply 341 by the units digit of 22, which is 2:
  • $341 \times 2 = 682$
• Step 3: Multiply 341 by the tens digit of 22, which is also 2, but remember to shift one position to the left because it is a tens place:
  • $341 \times 2 = 682$
  • Write this product as 6820 (shifted one position to the left).
• Step 4: Add the results from steps 2 and 3:
  • $682 + 6820 = 7502$

Therefore, the solution to the problem is $7502$.

To solve this problem, we'll follow these steps:

• Multiply $114$ by the unit digit of $13$, which is $3$.

• Multiply $114$ by the tens digit of $13$, which is $1$.

• Add the resulting products, considering place values.

Now, let's carry out the multiplication step-by-step:
Step 1: Multiply $114$ by $3$. This gives $114 \times 3 = 342$. Write this result as the first partial product.
Step 2: Multiply $114$ by $1$. This gives $114 \times 1 = 114$. Since this is a tens digit multiplication, place this result one position to the left, resulting in $1140$.

Step 3: Add the partial products:
$\begin{aligned} & & 342 \\ + & & 1140 \\ \hline & & 1482 \end{aligned}$

Therefore, the product $114 \times 13 = 1482$.

The correct answer is $1482$.

To solve this problem, we'll follow these steps:

• Step 1: Multiply $173$ by $3$, which is the first digit of $30$.
• Step 2: Consider the zero from $30$ to multiply the entire result by $10$.
• Step 3: Add the results to find the product.

Now, let's work through each step:

Step 1: Multiply $173 \times 3$.

$173 \times 3 = 519$.

Step 2: Multiply the result $519$ by $10$ since we're considering $30$.

$519 \times 10 = 5190$.

Therefore, the solution to the problem is $5190$.

To solve this problem, we'll perform vertical multiplication of $162$ by $13$ as follows:

• Step 1: Multiply $162$ by $3$ (the units digit of $13$): $\begin{aligned} \quad 162 \times 3 &= 486 \end{aligned}$

• Step 2: Multiply $162$ by $1$ (the tens digit of $13$), remembering that this $1$ is in the tens place: $\begin{aligned} \quad 162 \times 10 &= 1620 \end{aligned}$

• Step 3: Add the two results from Steps 1 and 2: $\begin{aligned} \quad 486 + 1620 &= 2106 \end{aligned}$

Therefore, the product of $162$ and $13$ is $2106$, which corresponds to choice (2).