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

To solve this problem, follow these steps:

• Step 1: Identify the given numbers: 72 and 20.
• Step 2: Use vertical multiplication to multiply these numbers.
• Step 3: Break down the multiplication into simpler steps and perform the calculations.

Let's work through each step:

Step 1: The given numbers are 72 and 20.

Step 2: We'll perform vertical multiplication by breaking down the problem into parts. Start with:

$72 \times 20$

Step 3: Multiply 72 by each digit of 20 separately.

• First, multiply 72 by 0 (the units digit of 20).
  $72 \times 0 = 0$.
• Next, multiply 72 by 2 (the tens digit of 20), noting it represents 20.
  $72 \times 2 = 144$.
  Since this 2 is in the tens place, we write this as 1440 (i.e., 144 shifted one place to the left).

Now add the results:
- From $72 \times 0$: 0
- From $72 \times 20$: 1440

Adding these results gives:
$0 + 1440 = 1440$.

Therefore, the solution to the problem is $1440$.

To solve the multiplication problem $29 \times 16$, we'll follow these steps using the vertical multiplication method:

• First, multiply the digit in the units place of 16 by 29: $6 \times 29$.
• Then, multiply the digit in the tens place of 16 (1) by 29, and shift one position to the left (because it represents a ten): $10 \times 29$.
• Add the two products to get the final product.

Let's work through each step:

Step 1: Calculate $6 \times 29$.
- $29 \times 6 = 174$.

Step 2: Calculate $10 \times 29$.
- $29 \times 1 = 29$, and then shift one position to the left to get $290$.

Step 3: Add both results.
- $174 + 290 = 464$.

Thus, the solution to $29 \times 16$ is $464$.

To find the value of $x$ using vertical multiplication, follow these steps:

• Write the two numbers vertically, with 25 on top and 13 below it:

• Multiply the digit in the units place of 13 (which is 3) by each digit in 25:
• 3 times 5 equals 15. Write 5 in the units place and carry over 1
• 3 times 2 equals 6, plus the carried over 1 equals 7
$\begin{array}{c} \ \ \ \ 75 \end{array}$
• Next, multiply the digit in the tens place of 13 (which is 1) by each digit in 25. Remember to shift this product one place to the left:
• Starting beneath the tens column, 1 times 5 equals 5
• 1 times 2 equals 2
$\begin{array}{c} \ \ 250 \end{array}$
• Add the two partial products together to get the final product:
$\begin{array}{c} \ \ \ 75 \\ + \ 250 \\ \hline \ 325 \end{array}$

Therefore, the value of $x$ is $325$.

To solve this problem, we'll perform the following steps using column multiplication:

• Step 1: Write the numbers in a column, with 64 on top and 16 beneath, ensuring proper alignment.
• Step 2: Multiply the units digit (6) of the bottom number (16) by the entire top number (64).
• Step 3: Multiply the tens digit (1) of the bottom number (16) by the entire top number (64) and shift the result one place to the left (equivalent to multiplying the result by 10).
• Step 4: Add these two partial products to find the final result.

Let's work through this step-by-step:

Step 1: Set up the multiplication vertically.

$\begin{array}{c} \phantom{+}64 \\ \times 16 \\ \end{array}$

Step 2: Multiply the units digit of the bottom number (6) by the entire top number (64):
$6 \times 64 = 384$.

Step 3: Multiply the tens digit of the bottom number (1) by the entire top number (64) and remember to place a zero since this is in the tens place:
$1 \times 64 = 64$, shift by one place to get $640$.

Step 4: Add the two partial products:
$384 + 640 = 1024$.

Therefore, the solution to the problem is $1024$.

To solve this problem, we will perform vertical multiplication of the numbers $73$ and $13$.

First, write the numbers vertically:

$73$

$\times \ 13$

Step 1: Multiply the units digit of $13$ (which is $3$) by each digit of $73$:

• $3 \times 3 = 9$
• $3 \times 7 = 21$

The partial result is $219$.

Step 2: Multiply the tens digit of $13$ (which is $1$, but treat it as $10$) by each digit of $73$:

• $10 \times 3 = 30$
• $10 \times 7 = 70$

The partial result is $730$ because we start from the tens place.

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

$\begin{array}{c} \ \ \ \,219\\ + \ 730\\ \hline \ 949\\ \end{array}$

Therefore, the product of $73$ and $13$ is $949$.

The solution is $\boxed{949}$, which corresponds to choice $4$.

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

• Step 1: Write the numbers vertically, aligning by place value.
• Step 2: Multiply the digit in the units place of the lower number by the entire upper number.
• Step 3: Multiply the digit in the tens place of the lower number by the entire upper number and shift one place to the left.
• Step 4: Add the products obtained in Step 2 and Step 3.

Let's work through each step:
Step 1: Arrange the numbers:

• ...... 36
• x .... 11

Step 2: Multiply the upper number by the units digit of the lower number (1):

• 36 (since $36 \times 1 = 36$)

Step 3: Multiply the upper number by the tens digit of the lower number (1) and remember to shift by one place (which represents multiplying by 10):

• 360 (since $36 \times 10 = 360$)

Step 4: Add the results from Steps 2 and 3:

• .......... 36
• + ......360
• -----------
• ).......396

Therefore, the product of 36 and 11 is $396$, which corresponds to choice 2.

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

• Step 1: Identify the numbers to be multiplied, which are 20 and 21.
• Step 2: Multiply 20 by 21 using simple arithmetic rules for multiplication.
• Step 3: Compare the result to the provided answer choices to confirm the correct one.

Now, let's work through each step:
Step 1: The numbers given are 20 and 21.
Step 2: We'll use the formula $a \times b = \text{Product}$, which becomes $20 \times 21$.
To simplify the multiplication:
- Multiply each digit: $20 \times 20 = 400$ and $20 \times 1 = 20$.
- Add the products together: $400 + 20 = 420$.

Therefore, the solution to the problem is $x = 420$.

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

• Step 1: Set up the numbers in vertical format for multiplication.
• Step 2: Multiply each digit of the number 16 by each digit of the number 11, starting with the rightmost digit.
• Step 3: Sum the results, considering the appropriate place values.

Now, let's proceed with the steps in detail:

Step 1: Write the numbers vertically:
           16
×        11
--------------------

Step 2: Multiply as follows:
- Multiply 16 by 1 (the units place of the lower number):
              16

- Multiply 16 by 10 (the tens place of the lower number, written as 1 * 10):
          160   (shifted one place to the left)

Step 3: Add both results. Here, aligning by place values is crucial:

            16
+        160
--------------------
          176

Therefore, the product of 16 and 11 is $176$.

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

• First, multiply the ones digit of 19 by 32: $9 \times 32$.
  $9 \times 2 = 18$ (write 8, carry over 1).
  $9 \times 3 = 27$. Add the carry over: $27 + 1 = 28$.
  This gives us the partial product 288.
• Next, multiply the tens digit of 19 by 32, remembering to align correctly by adding a zero at the end: $1 \times 32$ (shift result one position to the left).
  $1 \times 2 = 2$ and $1 \times 3 = 3$, thus giving 32.
  Write this as 320 (by placing a 0 after, to account for tens place).
• Add these two partial products:
    288
  + 320
  -------
    608

Therefore, the product of 32 and 19 is $608$.

To find $x$ in this multiplication problem, we will multiply the numbers 31 and 17 step-by-step:

Write down the numbers as they would appear in vertical multiplication:

    31
  × 17
  -----

Step 1: Multiply the units digit of 17, which is 7, by the entire number 31.

• $31 \times 7 = 217$

Write this partial product below the line:

    31
  × 17
  -----
   217

Step 2: Multiply the tens digit of 17, which is 1, by the entire number 31, remembering this is actually multiplying by 10, so we add a zero at the end.

• $31 \times 1 = 31$. Since we are multiplying by 10, it becomes $310$.

Write this partial product, shifted one position to the left (adding a zero):

    31
  × 17
  -----
   217
+ 310
  -----

Step 3: Add the two partial products together to get the final product.

• Adding $217$ and $310$ gives us $527$.

Thus, the product of 31 and 17, which is the value of $x$, is $527$.

To solve this problem, we'll perform vertical multiplication between the numbers 26 and 13. Here are the steps:

• Step 1: Multiply the digits in the units place.
• Step 2: Multiply the units place digit of the second number by the first number:

$3 \times 26$

Calculate $6 \times 3 = 18$. Write down 8, carry over 1.

Then $2 \times 3 = 6$ plus the carry over 1 gives 7.

Result: 78.

• Step 3: Multiply the digit in the tens place of the second number by the first number.

$1 \times 26 = 26$. Since this is tens place, write down 260 (shift one place to the left).

• Step 4: Add the two results.

$78 + 260 = 338$

Therefore, the product of 26 and 13 is $\textbf{338}$.

To solve this problem, we'll employ the vertical multiplication method to compute $27 \times 21$.

• First, multiply $27$ by the units digit of $21$ (which is $1$):
  $27 \times 1 = 27$
• Next, multiply $27$ by the tens digit of $21$ (which is $2$) and shift the result one place to the left:
  $27 \times 2 = 54$. Because this represents tens, it becomes $540$.
• Sum the two products obtained above:
  $27 + 540 = 567$

Therefore, the solution to the problem is $567$.

To solve this problem, we'll multiply the two-digit numbers 21 and 15 using vertical multiplication:

• Step 1: Multiply the ones digit of 15 (5) by each digit of 21.

  - $5 \times 1 = 5$
  - $5 \times 2 = 10$

  So, we write this partial product as 105, ensuring that we account for place value.

• Step 2: Multiply the tens digit of 15 (1, which is actually 10) by each digit of 21, adding one zero beforehand for the tens place.

  - $1 \times 1 = 1$ (Since 1 is actually 10, consider this as $10 \times 1 = 10$)
  - $1 \times 2 = 2$ (Since 1 is actually 10, consider this as $10 \times 2 = 20$)

  Write this second partial product as 210, remembering the zero addition for the tens place.

• Step 3: Add the partial products from Step 1 and Step 2.

  $105 + 210 = 315$

So, the final product of 21 and 15 is $315$.

We can confirm that the correct choice among the provided options is $315$.

To solve the problem of finding $x$, we will calculate the area of a rectangle with sides 25 and 15. This requires multiplying these two lengths:

Here is a step-by-step breakdown of the multiplication process using the column method:

• Step 1: Multiply the unit value of 15 by 25:
  $5 \times 25 = 125$
• Step 2: Multiply the tens value of 15 (which is 10) by 25:
  $10 \times 25 = 250$
• Step 3: Add the results of step 1 and step 2:
  $125 + 250 = 375$

Therefore, the value of $x$ is the area of the rectangle, which equals $375$.

The correct choice corresponding to this result is option 2.

Thus, the solution to the problem is $x = 375$.

To solve this problem, we will multiply the numbers 92 and 10. Here's how:

• Step 1: Write the problem vertically with 92 on top and 10 below it because 10 is a simpler number to multiply.
• Step 2: Start by multiplying 92 by 0, which gives 0.
• Step 3: Next, multiply 92 by 1 (from 10) but remember this is in the tens place, so it actually represents multiplying 92 by 10, resulting in 920.

Multiplication Breakdown:
92
× 10
------
920

Therefore, 92 multiplied by 10 equals $920$.

Thus, the solution to the problem is $920$.

To solve the problem of multiplying 16 by 12 using vertical multiplication, follow these steps:

• Step 1: Multiply the units digit of 12 (2) by 16.
• Step 2: $2 \times 6 = 12$. Write 2, carry 1.
• Step 3: $2 \times 1 + 1 = 3$. So, the result is 32.
• Step 4: Multiply the tens digit of 12 (1) by 16.
• Step 5: $1 \times 16 = 16$. Shift this result one place to the left (add a trailing zero).
• Step 6: Add the partial products: 32 + 160 = 192.

Therefore, the product of $16 \times 12$ is $192$.

To solve this problem, we will multiply 33 by 17 using vertical multiplication:

• Step 1: Multiply 33 by the unit digit of 17, which is 7.
• Step 2: Multiply 33 by the tens digit of 17, which is 1, and remember to position its result correctly as it's equivalent to multiplying by 10.
• Step 3: Sum the results obtained in the above steps to get the final product.

Let's perform the calculations:

Step 1: $33 \times 7 = 231$

Step 2: $33 \times 10 = 330$ (since the tens digit of 17 is 1, effectively $33 \times 1 \times 10$)

Step 3: Add these two results:

$231 + 330 = 561$

Therefore, the solution to this problem is $561$.

To solve the vertical multiplication problem of $26 \times 11$, we proceed as follows:

• First, write 26 and 11 as a vertical multiplication setup.
• Step 1: Multiply 26 by the digit in the units place of 11, which is 1:
$26 \times 1 = 26$
• Step 2: Multiply 26 by the digit in the tens place of 11 (when thinking about multiplication, consider it 10):
$26 \times 10 = 260$
• Ensure to shift the result by one place to the left since we are multiplying by the tens place;
• Add the two results together:
$26 + 260 = 286$

Therefore, the product of $26 \times 11$ is $286$.

To solve this problem, we will multiply the two numbers 25 and 16 using the standard method of vertical multiplication. Let's break this down step-by-step:

First, multiply the smallest place value:

• Multiply the units digit of the second number (6) by the first number (25):
  $25 \times 6$:
• Multiply: $6 \times 5 = 30$, write 0 and carry over 3.
• Multiply: $6 \times 2 = 12$, add the carryover 3 to get 15.
  This gives us 150.

Next, multiply the tens digit of the second number (1) by the first number (25), remembering to shift this result one place to the left (equivalent to multiplying by 10):

• $25 \times 10 = 250$
• The multiplication by 1 here doesn't change the product, only the shift does.

Now, add the partial products obtained from the above steps:

• Add: $150 + 250$
• Result: $400$

Therefore, the solution to the problem is $400$.

To solve this problem, we will use vertical multiplication to find the product of 82 and 30. Begin with these steps:

• Step 1: Write the numbers in vertical format:

      82
    x 30
    ----
    

• Step 2: Start by multiplying the 2 in the unit's place of 82 by each digit of 30. Since 30 is composed of 0 and 3 (which represents 30 because it’s in the tens place), the small segments of multiplication become:

  • $82 \times 0 = 0$

  • Record this result beneath the line, shifted to match the units place. Since this is multiplication by zero, this line remains:

            0
         

• Step 3: Now multiply the next digit in the number 30: $\text{Since it is } 3 \times 82,\text{ and it is in the tens place, it actually represents } 30 \times 82.$

  • First, multiply $82 \times 3 = 246$

  • Then multiply 246 by 10 (since 3 is actually 30):

         2460
         

• Step 4: Add the two results (noting the multipliers from each digit of 30 already included their place value):

      82
    x 30
    ----
      0   (from 82 x 0)
    2460  (from 82 x 30)
    ----
    2460
    

• Therefore, the product of $82 \times 30$ is $\mathbf{2460}$.

Therefore, the correct answer to the problem is $2460$, which corresponds to choice (1).