Vertical Multiplication - Examples, Exercises and Solutions

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

• Step 1: Multiply the unit digits.
• Step 2: Multiply the tens digits.
• Step 3: Add the results from Steps 1 and 2.

Let's execute these steps:
Step 1: Multiply the unit digit of 53, which is 3, by 3:
$3 \times 3 = 9$.

Step 2: Multiply the tens digit of 53, which is 5 (standing for 50), by 3:
$50 \times 3 = 150$.

Step 3: Add the results of Step 1 and Step 2:
$150 + 9 = 159$.

Therefore, the solution to the problem is $159$.

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 $248$.

We will solve the problem using direct multiplication of the two numbers, 30 and 4.

Steps:

• First, multiply the one's place of 30 by 4:
  $0 \times 4 = 0$

• Second, multiply the ten's place of 30 by 4:
  $3 \times 4 = 12$

• The result from the tens multiplication is over the magnitude of the number 30 (since it's in the tens place), so we already account for place by multiplying 3 by 4 directly forming a product 12, no tens digit carries from one's digit.

• Combine these results to get the total product:
  $0 + 120 = 120$

Therefore, the product of $30$ and $4$ is $\mathbf{120}$.

By referencing the multiple-choice options provided, the correct choice matches the calculation we performed and is choice 3: $120$.

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 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 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 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 follow these steps:

• Step 1: Multiply the ones digit of the two-digit number by the single-digit number.
• Step 2: Multiply the tens digit of the two-digit number by the single-digit number.
• Step 3: Add the two products from the above steps to find the final result.

Now, let's work through each step:
Step 1: Multiply $5 \times 9$. This results in $45$, which includes the 5 in the ones place, and we carry over 4.
Step 2: Next, multiply $2 \times 9$ (from the tens place), which equals $18$. Add the carried-over 4 to get $18 + 4 = 22$. This 22 represents 220 when taking place value into account.
Step 3: Combining steps 1 and 2, we put the $5$ from step 1 in the ones digit and the result from step 2 as tens (which corresponds to $220 + 5 = 225$).

Therefore, the solution to the problem is $x = 225$, aligning with choice 4.

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 problem, we'll multiply $42$ by $7$ using vertical multiplication.

• Step 1: Break down $42$ as $40 + 2$.
• Step 2: Multiply each part by $7$.

Let's perform the calculations:

• Multiply $2 \times 7$ makes $14$.
• Multiply $40 \times 7$ equals $280$.
• Add the two products: $280 + 14 = 294$.

Therefore, the solution to the multiplication problem is $\boxed{294}$.

Upon reviewing the provided choices, the correct choice is option $3$ with result $294$.

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

• Step 1: Identify the given numbers. We have $45$ and $8$.
• Step 2: Perform vertical multiplication of $45 \times 8$.

Now, let's work through each step:

• Multiply the units digit of 45 by 8:
  $5 \times 8 = 40$.
  Write 0 in the units place and carry over 4 to the tens.
• Multiply the tens digit of 45 by 8, and add the carry-over:
  $4 \times 8 = 32$.
  Add the carry-over 4: $32 + 4 = 36$.
• Write 36 in the tens and hundreds place, giving us the final product:

Combining these, the final result of the multiplication is $360$.

Therefore, the solution to the problem is $360$, which corresponds to choice number 3.

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 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 multiplication problem, follow these clear steps:

• Step 1: Align the numbers vertically (place 26 above 6), ensuring the digits are properly arranged by place value.
• Step 2: Begin multiplication with the unit digit of the bottom number (6). Multiply 6 by each digit in 26, starting from the right.

Now, let's perform the calculations:

Step 1: Multiply the units digit of 6 with the number 26:
- $6 \times 6 = 36$. Write 6 in the units place of the answer, and carry over the 3.
- Next, multiply $6 \times 2 = 12$. Then, add the carryover (3) to 12, resulting in 15.

Step 2: Write 15 next to the 6 in the result. Thus, the complete multiplication gives 156.

Therefore, the solution to the problem is $\boxed{156}$.

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

• Step 1: Identify the given information
• Step 2: Perform the multiplication of 36 by 5
• Step 3: Verify the product against the provided answer choices

Now, let's work through each step:
Step 1: We are given the numbers 36 (a two-digit number) and 5 (a single-digit number).
Step 2: Perform direct multiplication:
Multiply the units digit of 36 by 5: $6 \times 5 = 30$. Write down the 0 and carry over the 3.
Multiply the tens digit of 36 by 5: $3 \times 5 = 15$. Add the carry-over 3 to get 18.
Combine these results to form the full product: 180.
Step 3: The calculated product is 180. Comparing this with the provided answer choices, the correct choice is $180$.

Therefore, the solution to the problem is $180$.