Vertical Multiplication - Examples, Exercises and Solutions

To solve this problem, we will multiply 24 by 7 using standard multiplication:

• Step 1: Multiply the unit digit of 24 by 7:
  $4 \times 7 = 28$. Write down 8 and carry over 2.
• Step 2: Multiply the tens digit of 24 by 7, then add the carry over:
  $2 \times 7 = 14$, and add the carried-over 2 to get 16.

The final result of these calculations is:
Since the unit's place is 8 and the ten's place is 16, our final answer is $168$.

Therefore, the solution to the problem is $168$.

To solve this problem, we'll follow these steps to multiply $77$ by $3$:

First, set up the numbers for vertical multiplication:

$\begin{array}{c} & 77 \\ \times & \,\,3 \\ \hline \end{array}$

• Step 1: Multiply the units:

$7 \times 3 = 21$

Write down $1$ and carry over $2$.

• Step 2: Multiply the tens:

$7 \times 3 = 21$

Add the carry-over $2$, resulting in $23$.

Write down $23$ as there are no more digits to multiply.

Combining both steps, we find the product of $77$ and $3$ is:

$\begin{array}{c} & 77 \\ \times & \,\,3 \\ \hline & 231 \\ \end{array}$

Therefore, the solution to the problem is $231$.

To solve this problem, we'll employ vertical multiplication.

Step 1: Set up the multiplication:
        $62$
×       $4$
      ---------

Step 2: Multiply each digit of 62 by 4. We start with the ones place, then the tens place.

• Multiply the ones digit: $2 \times 4 = 8$.
• Multiply the tens digit: $6 \times 4 = 24$.

Step 3: Consider the place value for each part of the calculation:
The result from multiplying the tens digit by 4 represents $240$ because it is $24 \times 10$.

Step 4: Add the two partial results:
         8
+ 240
      ---------
      248

Therefore, the solution to the problem is $248$.

To solve this problem, we will multiply $91$ by $3$ using standard multiplication techniques:

• Step 1: Multiply the unit digit of $91$ by $3$:
  $1 \times 3 = 3$.
• Step 2: Multiply the tens digit of $91$ by $3$:
  $9 \times 3 = 27$.
• Step 3: Place the result of $27$ correctly one digit to the left (because it's actually $90 \times 3$), which gives $270$.
• Step 4: Add the results from Step 1 and Step 3:
  $270 + 3 = 273$.

Therefore, the product of $91 \times 3$ is $273$.

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

• Step 1: Identify the given numbers: $86$ and $3$.
• Step 2: Perform the vertical multiplication of $86$ by $3$.

Let's work through each step:

Step 1: We will multiply $86$ by $3$. The multiplication can be broken down as follows:

1. Multiply the units digit of $86$, which is $6$, by $3$:

$6 \times 3 = 18$

Write down $8$ and carry over $1$ to the next column (the tens place).

2. Multiply the tens digit of $86$, which is $8$, by $3$:

$8 \times 3 = 24$

Add the carried over $1$ to $24$:

$24 + 1 = 25$

Write down $25$. Since we're only multiplying a two-digit number by a one-digit number, our result directly follows:

Therefore, the solution to the problem is $258$.

To solve this problem, we'll perform a vertical multiplication of the two-digit number 82 by the one-digit number 9.

• Step 1: Write down the numbers:
                 82
         $\times$ 9
          ---
• Step 2: Multiply each digit of 82 by 9 starting from the unit's digit:
      $2 \times 9 = 18$
      Write 8 in the unit's place and carry over 1.
• Step 3: Multiply the tens digit, taking into account the carried over value:
      $8 \times 9 = 72$
      Add the carried over 1: $72 + 1 = 73$
• Step 4: Write down the result:
                 738

Therefore, the product of the multiplication is $\mathbf{ 738 }$.

By comparing this result with the provided options, option 2 is the correct solution.

To solve this multiplication problem, we will perform the following steps:

• Step 1: Write the two-digit number 19 as the sum of its place values: 19 = 10 + 9.
• Step 2: Multiply the first term (10) by 6: $10 \times 6 = 60$.
• Step 3: Multiply the second term (9) by 6: $9 \times 6 = 54$.
• Step 4: Add the results of the two multiplications: $60 + 54 = 114$.

Therefore, the product of 19 and 6 is $\textbf{114}$.

To solve this problem, we'll start from the equation that needs to be true:

$74x = 592$

We want to solve for $x$. To do this, we divide both sides of the equation by 74:

$x = \frac{592}{74}$

Now, we'll perform the division:

$\frac{592}{74} = 8$

This tells us that:

$x = 8$

By substituting back into the multiplication:

$74 \times 8 = 592$

The calculation is verified. Therefore, the solution to the problem is:

$592$

Given the multiple-choice options, option 2 corresponds to this solution:

The correct answer is $x = 8$, confirming choice 2.

To solve this multiplication problem, we will use the vertical multiplication method:

• Step 1: Multiply the ones digit of the first number by the second number.
• Here, multiply $3 \times 7 = 21$. Record the 1 in the ones place and carry over the 2.
• Step 2: Multiply the tens digit of the first number by the second number, and add any carried over value from the first step.
• Calculate $7 \times 7 = 49$. Add the carry-over of 2 to this result, which gives $49 + 2 = 51$.
• Write the 51 on top of where we placed our previous result, so it becomes 5 at the tens and hundreds position.

Therefore, the final multiplied value is $511$.

The correct answer choice is option 4: $511$.

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

• Step 1: Multiply each digit of the number $92$ by $7$.
• Step 2: Combine results considering their place values.
• Step 3: Sum these partial products for the final answer.

Now, let's work through each step:

Step 1: Multiply the units digit of $92$, which is $2$, by $7$:
$2 \times 7 = 14$. Record $4$ in the units place and carry over $1$.

Step 2: Multiply the tens digit of $92$, which is $9$, by $7$:
$9 \times 7 = 63$. Add the carry-over $1$ to get $64$.

Step 3: Record the $4$ from $64$ in the tens place and place $6$ in the hundreds place.
Thus, arranging our final result: $644$.

Therefore, the product of $92$ and $7$ is $644$.

To solve this problem, we'll multiply the numbers directly:

• Step 1: Identify the numbers to multiply: $32$ and $8$.
• Step 2: Use the vertical multiplication method to calculate the product.
• Step 3: Verify the calculation by using the distributive property as a secondary method.

Now, let's work through each step:
Step 1: We have the multiplicand $32$ and the multiplier $8$.
Step 2: Multiply $32$ by $8$. To do this, break it down as follows:
- $32 = 30 + 2$
- Multiply each part by 8: $30 \times 8 = 240$ and $2 \times 8 = 16$.
- Add the two products together: $240 + 16 = 256$.

Step 3: Verify this by rechecking the arithmetic or using properties of multiplication.

Therefore, the solution to the problem is $256$.

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

• Step 1: Breakdown the multiplication into tens and units.
• Step 2: Multiply each digit by the single-digit multiplier.
• Step 3: Sum the partial results to get the final product.

Let's work through each step:

Step 1: We need to multiply each digit of the number 16 by 5.
- The tens digit of 16 is 1 (representing 10), and the units digit is 6.

Step 2: Perform the multiplication:
- Multiply the tens digit: $10 \times 5 = 50$
- Multiply the units digit: $6 \times 5 = 30$

Step 3: Add the results from these multiplications:
- Total: $50 + 30 = 80$

Therefore, the solution to the problem is $80$.

To solve the multiplication problem $64 \times 6$, we'll perform the following steps:

• Step 1: Break down $64$ into tens and units. So, $64$ can be written as $60 + 4$.
• Step 2: Multiply each component separately by $6$.
• Step 3: Calculate $60 \times 6$ and $4 \times 6$ separately.
• Step 4: Sum the results of the above calculations to find the total product.

Now, let's execute these steps specifically:

Step 1: Represent $64$ as $60 + 4$. This simplifies the multiplication process.

Step 2: Multiply the tens: $60 \times 6 = 360$.

Step 3: Multiply the units: $4 \times 6 = 24$.

Step 4: Now, add the two results: $360 + 24 = 384$.

Therefore, the product of $64 \times 6$ is $384$.

To solve this problem, we'll perform vertical multiplication of $82$ by $2$:

• Step 1: Multiply the ones place. Multiply $2$ (from 82) by $2$:

$2 \times 2 = 4$
This gives us $4$ in the ones place.

• Step 2: Multiply the tens place. Multiply $8$ (in the tens place of 82) by $2$:

$8 \times 2 = 16$
Since the result is $16$, we place $6$ in the tens place and carry over $1$ to the next higher place (hundreds place).

• Step 3: Add up the intermediate results.

The ones place has $4$, and the tens place has $6$ plus $1$ (carry-over), totaling to $7$ in the tens place. Thus, the full number now reads:

$164$

Therefore, the solution to the problem is $164$.

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

• Step 1: Multiply the units digit of 73 by 8.
• Step 2: Multiply the tens digit of 73 by 8.
• Step 3: Add the results of these two multiplications.

Now, let's work through each step:

Step 1: Multiply the units digit of 73 (which is 3) by 8:
$3 \times 8 = 24$
We'll write 4 in the ones place of the result and carry over 2 to the tens place.

Step 2: Multiply the tens digit of 73 (which is 7) by 8 and add the carried over 2:
$7 \times 8 = 56$
Adding the carried over 2 gives us:
$56 + 2 = 58$

Step 3: Write the result from the tens multiplication in the tens and hundreds place:
Combining our results, we get:
$73 \times 8 = 584$

Therefore, the solution to the problem is $584$.