Examples with solutions for Vertical Multiplication: 3-digit by 2-digit multiplication

Exercise #1

56063x

Video Solution

Step-by-Step Solution

To solve this problem, we'll perform the multiplication of 560 by 63 using the vertical multiplication method, which involves two main calculations:

  • Step 1: Multiply 560 by the unit digit of 63, which is 3.
  • Step 2: Multiply 560 by the tens digit of 63, which is 6 (representing 60), and shift the result one place to the left.
  • Step 3: Sum the results from steps 1 and 2.

Now, let's perform the calculations:
Step 1: Multiply 560 by 3:

560×3=1680 560 \times 3 = 1680

Step 2: Multiply 560 by 60 (which is 10 times 6):
First, multiply 560 by 6:

560×6=3360 560 \times 6 = 3360

Since we're actually multiplying by 60, we append a zero to the product (equivalent to multiplying by 10):
3360×10=33600 3360 \times 10 = 33600

Step 3: Add the results from steps 1 and 2:
1680+33600=35280 1680 + 33600 = 35280

Therefore, the product of 560 and 63 is 35280 \mathbf{35280} .

Answer

35280 35280

Exercise #2

31010x

Video Solution

Step-by-Step Solution

To solve this problem, we'll multiply 310310 by 1010 through the following steps:

  • Step 1: Set up 310310 and 1010 for multiplication vertically:
  • 310×10 \begin{array}{c} 310 \\ \times \, 10 \\ \hline \end{array}
  • Step 2: Multiply each digit of 310310 by 1010:
    - Multiply the units digit of 310310 (which is 00) by 1010 gives 00.
    - Multiply the tens digit of 310310 (1) (1) by 1010: (1×10=10)(1 \times 10 = 10), shifted over two decimal places.
    - Multiply the hundreds digit of 310310 (3) (3) by 1010: (3×10=30)(3 \times 10 = 30), shifted over two decimal places.
  • 310×103100 \begin{array}{c} 310 \\ \times \, 10 \\ \hline 3100 \\ \end{array}
  • Step 3: Sum these products (effectively, it's already in 31003100 due to the design of the operation).

The solution is therefore 3100\mathbf{3100}.

Answer

3100 3100

Exercise #3

43011x

Video Solution

Step-by-Step Solution

To solve this problem, we will multiply 430 by 11 using vertical multiplication:

  • Step 1: Multiply 430 by 1 (the ones place of 11).
    430 multiplied by 1 is 430. Let's write this down while aligning according to place value, i.e., 430.

  • Step 2: Multiply 430 by 10 (the tens place of 11).
    To do this, multiply 430 by 1 and shift the result one place to the left, adding a zero at the end. Thus, 430 multiplied by 10 is 4300.

  • Step 3: Add the results from step 1 and step 2.
    Align the numbers:
            430
    + 4300
    --------
          4730

Therefore, the product of 430 and 11 is 4730\textbf{4730}.

The correct answer is 4730, which corresponds to choice 2.

Answer

4730 4730

Exercise #4

16112x

Video Solution

Step-by-Step Solution

To solve the multiplication problem of 161 and 12, follow these steps:

  • First, multiply 161 by the rightmost digit of 12, which is 2.
  • Since 2×161=322 2 \times 161 = 322 , we write down 322.
  • Next, multiply 161 by the leftmost digit of 12, which is 1. Notice that this actually means 10, so we multiply and shift the result one place to the left.
  • 1×161=161 1 \times 161 = 161 . Write 161 below 322, but shift one position to the left (or add a zero at the end as a placeholder).
  • The shifted result is 1610.
  • Add the intermediate results: 322 + 1610.
  • 322+1610=1932 322 + 1610 = 1932 .

Therefore, the product of 161 and 12 is 1932 1932 .

Answer

1932 1932

Exercise #5

15013x

Video Solution

Step-by-Step Solution

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

  • Step 1: Multiply 150 150 by the units digit of 13 13 , which is 3 3 .
  • Step 2: Multiply 150 150 by the tens digit of 13 13 , which is 1 1 , but remember this represents 10 10 .
  • Step 3: Sum the results of these two steps to get the total result.

Now, let's work through each step:
Step 1: Multiply 150 150 by 3 3 .
150×3=450 150 \times 3 = 450
Step 2: Multiply 150 150 by 1 1 (considering the tens place value, this is actually multiplying by 10 10 ).
150×10=1500 150 \times 10 = 1500
Step 3: Add the two products together.
450+1500=1950 450 + 1500 = 1950

Therefore, the solution to the problem is 1950 1950 .

Answer

1950 1950

Exercise #6

16213x

Video Solution

Step-by-Step Solution

To solve this problem, we'll perform vertical multiplication of 162162 by 1313 as follows:

  • Step 1: Multiply 162162 by 33 (the units digit of 1313): 162×3amp;=486 \begin{aligned} \quad 162 \times 3 &= 486 \end{aligned}

  • Step 2: Multiply 162162 by 11 (the tens digit of 1313), remembering that this 11 is in the tens place: 162×10amp;=1620 \begin{aligned} \quad 162 \times 10 &= 1620 \end{aligned}

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

Therefore, the product of 162162 and 1313 is 2106 2106 , which corresponds to choice (2).

Answer

2106 2106

Exercise #7

17330x

Video Solution

Step-by-Step Solution

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

  • Step 1: Multiply 173173 by 33, which is the first digit of 3030.
  • Step 2: Consider the zero from 3030 to multiply the entire result by 1010.
  • Step 3: Add the results to find the product.

Now, let's work through each step:

Step 1: Multiply 173×3173 \times 3.

173×3=519173 \times 3 = 519.

Step 2: Multiply the result 519519 by 1010 since we're considering 3030.

519×10=5190519 \times 10 = 5190.

Therefore, the solution to the problem is 51905190.

Answer

5190 5190

Exercise #8

11413x

Video Solution

Step-by-Step Solution

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

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

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

  • Add the resulting products, considering place values.

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


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

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

The correct answer is 1482 1482 .

Answer

1482 1482

Exercise #9

34122x

Video Solution

Step-by-Step Solution

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:
    • 341×2=682341 \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:
    • 341×2=682341 \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=7502682 + 6820 = 7502

Therefore, the solution to the problem is 7502 7502 .

Answer

7502 7502

Exercise #10

23121x

Video Solution

Step-by-Step Solution

Let's solve this multiplication problem step by step:

  • Step 1: Multiply 231 by the units digit of 21, which is 1. 231×1=231 231 \times 1 = 231
  • Step 2: Multiply 231 by the tens digit of 21, which is 2 (representing 20) 231×2=462 231 \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 231 + 4620 = 4851

Therefore, the solution to the problem is 4851 4851 .

Answer

4851 4851

Exercise #11

10225x

Video Solution

Step-by-Step Solution

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):
5×2=10 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:
5×0+1=1 5 \times 0 + 1 = 1 . Write down 1.

Finally, multiply 5 by 1 (the hundreds digit of 102):
5×1=5 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:
2×2=4 2 \times 2 = 4 . Write down 4.

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

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

  • Step 4: Add these two results together.

510+20402550\begin{array}{c} 510 \\ +2040 \\ \hline 2550 \\ \end{array}

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

Therefore, the solution to the problem is 2550 2550 .

Answer

2550 2550

Exercise #12

11610x

Video Solution

Step-by-Step Solution

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

Our problem states:

116×x=1160 116 \times x = 1160

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

x=1160116 x = \frac{1160}{116}

Calculating the division, we find:

x=10 x = 10

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

Answer

1160 1160

Exercise #13

14216x

Video Solution

Step-by-Step Solution

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:
    142×6=852 142 \times 6 = 852

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

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

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

Therefore, the solution to the problem is 2272 2272 .

Answer

2272 2272

Exercise #14

12617x

Video Solution

Step-by-Step Solution

To solve the multiplication of 126126 and 1717, we will use vertical multiplication:

  • Write 126126 above 1717, 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 126126 by 77 (the ones digit of 1717): 126×7: 126 \times 7:
    126
    ×  7
    -----
     882
    
    This yields 882882.
  • Next, multiply 126126 by 11 (tens digit of 1717) with proper adjustment (shift one position to the left): 126×10: 126 \times 10:
    126
    ×10
    -----
    1260
    
    This results in 12601260.
  • Sum both intermediate products: 882+1260=2142 882 + 1260 = 2142

Therefore, the final product of 126×17126 \times 17 is 21422142.

Answer

2142 2142

Exercise #15

13111x

Video Solution

Step-by-Step Solution

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

  • First, multiply 131 by the units digit of 11 (1):
  • 131×1=131131 \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:
  • 131×10=1310131 \times 10 = 1310
  • Finally, sum the two products (considering the place values correctly):
  • 131+1310=1441131 + 1310 = 1441

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

Answer

1441 1441

Exercise #16

19120x

Video Solution

Step-by-Step Solution

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:
    191×0=0 191 \times 0 = 0
  • Step 2: Multiply 191 by 2 (the tens digit of 20).
    Let's set up the multiplication:
    191×2382 \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 38203820.
  • Step 4: Add the partial products:
    Since the first partial product was 0, we add:
    0+3820=3820 0 + 3820 = 3820

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

Answer

3820 3820

Exercise #17

13319x

Video Solution

Step-by-Step Solution

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 2527 .

Answer

2527 2527

Exercise #18

11170x

Video Solution

Step-by-Step Solution

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:

111 \quad\quad\quad\quad 111
×    70 \times\quad\quad\quad\; ~~70
  _____\quad\quad\quad\;\_\_\_\_\_
  7770 \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:

111×7=7770Place value adjustment777×10=7770 \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 x = 7770 .

Answer

7770 7770

Exercise #19

12031x

Video Solution

Step-by-Step Solution

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:

120×1=120120 \times 1 = 120

Step 2: Multiply 120 by 3:

120×3=360120 \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=3720120 + 3600 = 3720

Therefore, the product of 120120 and 3131 is 3720\textbf{3720}.

Answer

3720 3720

Exercise #20

13161x

Video Solution

Step-by-Step Solution

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

  • Step 1: Write the multiplication as follows:

+131×161 \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):

+131×16100131 \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:

+131×16107860+00131 \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:

+131×16107860+00131007991 \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 7991 .

This matches choice 3 in the given options.

Answer

7991 7991