Vertical Multiplication: Multiplying 2-Digit by 1-Digit Numbers

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 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 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 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 multiplication problem, we will use vertical multiplication:

• Step 1: Write down the multiplication in the vertical form:
         $96$
  $\times$     $3$
  $\_\_\_\_\_\_\_\_$
• Step 2: Multiply the one's digit of the bottom number (3) by the one's digit of the top number:\
  3 \times 6 = 18. Write 8 in the one's place and carry over 1 to the next place.
• Step 3: Multiply the tens digit of the top number (9) by 3:
  3 \times 9 = 27. Add the carryover 1, getting 28. Write 28 in the tens and hundreds places.
• Step 4: Write down the results:
       $288$

Therefore, the product of $96 \times 3$ is $288$.

Hence, the correct answer is choice $288$.

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'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 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 $180$.

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 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 problem, we'll follow these steps:

• Step 1: Identify the numbers to be multiplied: $15$ and $2$.
• Step 2: Calculate the product of these two numbers.

Now, let's work through each step:
Step 1: We are given the numbers $15$ and $2$.
Step 2: We will multiply these numbers together: $15 \times 2$ = $30$.

Therefore, the solution to the problem is $30$.

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 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 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 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 $584$.

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 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 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 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 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 $164$.