Vertical Multiplication: Multiplying 3-Digit by 1-Digit Numbers

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

• Step 1: Start by multiplying the number at the ones place of $152$ by $2$.
• Step 2: Move to the tens place of $152$ and multiply by $2$.
• Step 3: Finally, multiply the hundreds place of $152$ by $2$.
• Step 4: Sum these products, maintaining their respective place values to find the total product.

Let's perform the calculations:
Step 1: Multiply the ones digit:

$2 \times 2 = 4$

Step 2: Multiply the tens digit, and align it in the tens place:

$5 \times 2 = 10$.

Since $5 \times 2 = 10$, we write $0$ in the tens place and carry $1$ over to the hundreds place.

Step 3: Multiply the hundreds digit, and include the carry-over:

$1 \times 2 = 2$

Add the carry-over from the previous step:

$2 + 1 = 3$

Combine these results to form the final product:

The hundreds place is $3$, the tens place is $0$, and the ones place is $4$, resulting in:

$152 \times 2 = 304$

Therefore, the correct answer is $304$. Looking at the options given, the correct choice is option 2.

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

• Step 1: Align 163 for vertical multiplication by 3.
• Step 2: Multiply the units digit (3 from 163) by 3.
• Step 3: Multiply the tens digit (6 from 163) by 3, adding any carry from step 2.
• Step 4: Multiply the hundreds digit (1 from 163) by 3, adding any carry from step 3.

Now, let's work through each step:

Step 1: Multiply the units digit:
$3 \times 3 = 9$
Write 9 at the units place.
Carry over = 0 (since $9 < 10$).

Step 2: Multiply the tens digit:
$6 \times 3 = 18$
Add the previous carry (0):
Total = 18.
Write 8 at the tens place and carry over 1.

Step 3: Multiply the hundreds digit:
$1 \times 3 = 3$
Add the previous carry (1):
Total = 4.
Write 4 at the hundreds place.

The result of $163 \times 3$ is:

$489$

Therefore, the solution to the problem is $489$.

To solve this problem, let's perform the multiplication $171 \times 4$ using the vertical multiplication method:

• Step 1: Multiply the one's place: $4 \times 1 = 4$.
• Step 2: Multiply the ten's place: $4 \times 7 = 28$. Write 8 under ten's column and carry over 2.
• Step 3: Multiply the hundred's place: $4 \times 1 = 4$, then add the carry over: $4 + 2 = 6$.

Combine all parts to find the final result: The product is $684$.

Therefore, the solution to the problem is $684$.

To solve the problem of multiplying 175 by 4, we proceed as follows:

• Step 1: Set up the numbers for vertical multiplication, writing 175 on top and 4 beneath.
• Step 2: Multiply the rightmost digit of the top number, 5, by 4:
  • $5 \times 4 = 20$. Write 0 in the units place and carry over 2 to the next digit.
• Step 3: Multiply the next digit, 7, by 4 and add the carry over:
  • $7 \times 4 = 28$. Add the carry of 2 to this, resulting in 30. Write 0 in the tens place and carry over 3.
• Step 4: Multiply the leftmost digit, 1, by 4 and add the carry over:
  • $1 \times 4 = 4$. Add the carry of 3 to this, resulting in 7.
• Step 5: Combine the results from each step to form the complete product: 700.

Thus, the product of $175 \times 4$ is $700$.

Therefore, the correct answer is $700$, which corresponds to choice 3.

To solve the problem of multiplying 462 by 4, we will follow these steps:

• Write the numbers 462 and 4 in a vertical format:

462
×     4
────

• Step 1: Multiply the unit digit of 462 by 4.

$2 \times 4 = 8$.
Write 8 under the units column, as there is no carryover.

• Step 2: Multiply the tens digit of 462 by 4.

$6 \times 4 = 24$.
Write 4 under the tens column, and carry over 2 to the hundreds column.

• Step 3: Multiply the hundreds digit of 462 by 4.

$4 \times 4 = 16$.
Add the carryover 2: $16 + 2 = 18$.
Write 8 under the hundreds column and 1 in the thousands column.

The result will be read as follows from top to bottom in the columns:
1848

Therefore, the product of 462 and 4 is $1848$.

To solve the multiplication $341 \times 2$:

• Step 1: Multiply the units digit.
  $1 \times 2 = 2$. Write 2 in the units place.

• Step 2: Multiply the tens digit.
  $4 \times 2 = 8$. Write 8 in the tens place.

• Step 3: Multiply the hundreds digit.
  $3 \times 2 = 6$. Write 6 in the hundreds place.

Now, putting it all together, the product is $682$.

The solution to $341 \times 2$ is $682$.

To solve the problem of multiplying the numbers $130$ and $7$, we will perform vertical multiplication, focusing on each digit of the three-digit number:

• Step 1: Multiply the unit digit of $130$ by $7$.
  The unit digit of $130$ is $0$.
  $0 \times 7 = 0$ (carryover is 0).

• Step 2: Multiply the tens digit of $130$ by $7$.
  The tens digit of $130$ is $3$.
  $3 \times 7 = 21$ (write down $1$ and carry $2$ to the next column).

• Step 3: Multiply the hundreds digit of $130$ by $7$.
  The hundreds digit of $130$ is $1$. Including the carried over $2$, we calculate $1 \times 7 = 7$ plus $2$ equals $9$ (no further carryover is required).

Thus, the final step includes combining results: the hundreds place has $9$, tens place $1$, and the units place $0$, leading to the solution.

Therefore, the solution to the problem is $910$.

To solve this multiplication problem, we will use a step-by-step approach:

• Step 1: Write down the numbers in vertical format, aligning by place values:
                 $230$
              × $6$
• Step 2: Start multiplying from the rightmost digit:

First, multiply the units digit, $0 \times 6 = 0$.
Next, multiply the tens digit, $3 \times 6 = 18$. Write $8$ below the line in the tens place and carry over $1$ to the hundreds place.
Then, multiply the hundreds digit, $2 \times 6 = 12$. Add the carried $1$, giving $13$.

Thus, the multiplications line up to give $1380$ as the final product.

Therefore, the solution to the multiplication is $\boxed{1380}$.

To solve the multiplication problem $143 \times 4$, we'll follow these steps:

• Set up the numbers for vertical multiplication: write 143 and directly beneath it, write 4, followed by the multiplication line.
• Multiply each digit of 143 by 4, starting from the units place.
• Add any necessary carried numbers to the next multiplicative step.

Let's break down the multiplication:

• Step 1: Multiply the units place: $3 \times 4 = 12$. Write 2 in the units column, carry over 1 to the tens column.
• Step 2: Multiply the tens place: $4 \times 4 = 16$. Add the carried over 1, resulting in 17. Write 7 in the tens column and carry over 1 to the hundreds column.
• Step 3: Multiply the hundreds place: $1 \times 4 = 4$. Add the carried over 1, resulting in 5. Write 5 in the hundreds column.

By combining these, the result of $143 \times 4$ is $572$.

The choices provided are:

• Choice 1: $570$
• Choice 2: $572$
• Choice 3: $573$
• Choice 4: $574$

The correct answer corresponds to Choice 2: $572$.

Therefore, the solution to the problem is $572$.

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

• Step 1: Multiply each digit of 231 by 7.
• Step 2: Handle carry-over carefully.
• Step 3: Add results together to get the final product.

Let's execute the steps:
Step 1: Start from the rightmost digit of 231, which is the units place (1). Multiply it by 7:
$1 \times 7 = 7$

Step 2: Move to the tens digit, which is 3. Multiply it by 7 and add any carry-over. Since there was no carry from the first multiplication, we have:
$3 \times 7 = 21$
Write 1 in the tens place and carry over 2 to the hundreds place.

Step 3: Multiply the hundreds digit, which is 2. Multiply by 7 and add the carry-over of 2:
$2 \times 7 = 14$
Add the carry 2:
$14 + 2 = 16$

Write down the product from left to right: 16 (hundreds place), 1 (tens place), 7 (units place).
The full product is $1617$.

Therefore, the solution to the problem is $1617$.

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

• Step 1: Arrange the numbers for multiplication: write 921 on top and 3 below it with equal alignment.
• Step 2: Multiply the digit in the ones place of 921, which is 1, by 3.
• Step 3: Multiply the digit in the tens place of 921, which is 2, by 3.
• Step 4: Multiply the digit in the hundreds place of 921, which is 9, by 3.
• Step 5: Place each resulting product in the corresponding position in the final sum.
• Step 6: Sum all the individual results to get the final product.

Let's perform the calculations:

• Step 1: Multiply 1 by 3 giving us $3$ (put in the ones place).
• Step 2: Multiply 2 by 3 giving us $6$ (put in the tens place as 60).
• Step 3: Multiply 9 by 3 giving us $27$ (put in the hundreds place as 2700).
• Step 4: Sum up these results: $2700 + 60 + 3 = 2763$.

Therefore, the solution to the problem is $2763$.

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

• Step 1: Multiply the units digit of $341$ by $6$.
  The units digit is $1$, so $1 \times 6 = 6$.
• Step 2: Multiply the tens digit of $341$ by $6$.
  The tens digit is $4$, so $4 \times 6 = 24$.
  As $24$ is a two-digit number, place $4$ in the tens column and carry over $2$ to the hundreds column.
• Step 3: Multiply the hundreds digit of $341$ by $6$.
  The hundreds digit is $3$, so $3 \times 6 = 18$.
  Add the carried-over $2$ from the previous step to get $18 + 2 = 20$.

Finally, combining all the partial products, we get $2046$:

• Units column: $6$
• Tens column: $4$
• Hundreds column: $0$
• Thousands column (result of any carried-over sums): $2$

Therefore, the multiplication of $341$ by $6$ gives us the product $2046$.

To solve the multiplication problem $531 \times 5$, we will perform vertical multiplication, carefully managing place values and carries.

Let's break down the process:

• Step 1: Multiply the units digit (1) by 5:
  $1 \times 5 = 5$. Since this is a single-digit number, no carry is involved. The units place in the product is $5$.
• Step 2: Multiply the tens digit (3) by 5:
  $3 \times 5 = 15$. We place 5 in the tens place of the product and carry over 1 to the hundreds place.
• Step 3: Multiply the hundreds digit (5) by 5:
  $5 \times 5 = 25$. Add the carry of 1 from the previous step: $25 + 1 = 26$. We put the entire value 26 into the hundreds place and the thousands place as no more carries need to propagate beyond this point.

Constructing the final product from right to left, we have the value $2655$.

Therefore, the final solution to the multiplication problem $531 \times 5$ is $2655$.

To solve the problem of multiplying a three-digit number $451$ by a one-digit number $4$, let's use the vertical multiplication method.

Here are the steps:

• Step 1: Multiply the ones digit of $451$, which is $1$, by $4$.
  This gives us $1 \times 4 = 4$.
• Step 2: Multiply the tens digit of $451$, which is $5$, by $4$.
  Calculating $5 \times 4 = 20$.
  Remember to place this value one position to the left (as it represents tens, not units).
• Step 3: Multiply the hundreds digit of $451$ which is $4$, by $4$.
  This gives $4 \times 4 = 16$.
  This value represents hundreds, so its place value is shifted two positions to the left.

Now, let's put together these individual calculations for the final sum:
- For units: $4$
- For tens: $20$, but considered as $200$ in terms of place value.
- For hundreds: $1600$
The total sum is $1600 + 200 + 4 = 1804$.

Thus, the product of $451 \times 4$ is $\mathbf{1804}$.

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

• Step 1: Align the given numbers for multiplication.
• Step 2: Perform the multiplication digit by digit.
• Step 3: Sum the partial results.

Now, let's work through each step:
Step 1: The problem asks us to multiply $530$ by $6$. Set the numbers up such that $530$ is the multiplicand and $6$ is the multiplier.

Step 2: Let's multiply $6$ by each digit of $530$, from right to left:

• $6 \times 0 = 0$ (units place)
• $6 \times 3 = 18$. Write $8$ in the tens place and carry over $1$ to the hundreds place.
• $6 \times 5 = 30$. Adding the carried over $1$ gives $31$.

Step 3: Write down the resulting multiplication: $3180$.

Therefore, the solution to the multiplication problem is $3180$.

We will perform the multiplication of the numbers 146 and 7 using the vertical multiplication method:

• Step 1: Write the numbers vertically aligning the digits by place value:

$\phantom{999}146$
$\times \phantom{99}7$
$\underline{\phantom{999999999}}$

• Step 2: Multiply 7 by each digit of 146 starting from the right:

Multiply 7 by 6 (units place):

$7 \times 6 = 42$

• Write 2 in the units place and carry over 4 to the tens place.

Multiply 7 by 4 (tens place), add the carry over:

$7 \times 4 = 28$
$28 + 4 = 32$

• Write 2 in the tens place and carry over 3 to the hundreds place.

Multiply 7 by 1 (hundreds place), add the carry over:

$7 \times 1 = 7$
$7 + 3 = 10$

• Write 0 in the hundreds place and 1 in the thousands place.

Combine these to write out the full product:

$1022$

Therefore, the solution to the problem is $1022$.

Let us now perform the multiplication step-by-step:

• Step 1: Multiply the ones digit: $9 \times 9 = 81$. Write down $1$ and carry over $8$.
• Step 2: Multiply the tens digit: $3 \times 9 = 27$, then add the carryover $8 = 35$. Write down $5$ and carry over $3$.
• Step 3: Multiply the hundreds digit: $2 \times 9 = 18$, then add the carryover $3 = 21$. Write down $21$.
• Final Product: Arrange the results to form the number: $2151$.

Therefore, the product of $239$ and $9$ is $2151$.

Thus, the correct answer is $2151$.

To solve the multiplication problem $631 \times 6$ using vertical multiplication, follow these steps:

• Step 1: Begin with the rightmost digit of 631, which is 1.
• Multiply $1 \times 6 = 6$. Write 6 under the line.
• Step 2: Move to the next digit to the left, which is 3.
• Multiply $3 \times 6 = 18$. Write 8 below and carry over 1 to the next column.
• Step 3: Multiply the leftmost digit 6 by 6.
• Multiply $6 \times 6 = 36$. Add the carry-over 1 to get 37.
• Write down 37 on the left side.

The sum of these values is 3786.

Thus, the product of $631$ and $6$ is $\textbf{3786}$.

The correct answer, as per the answer choices, is 3786.

To solve this problem, we will perform the multiplication step by step:

• Step 1: Multiply the units digit of 720 by 6, which is $0 \times 6 = 0$.
• Step 2: Multiply the tens digit of 720 by 6 and add any carryovers. So, $2 \times 6 = 12$. Write down 2 and carry over 1.
• Step 3: Multiply the hundreds digit of 720 by 6 and add any carryovers. So, $7 \times 6 = 42$, and add the 1 carried over, resulting in 43.

Combining these results gives us 4320. Thus, the product of 720 and 6 is $4320$.

Therefore, the solution to the problem is $4320$.

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

• Step 1: Multiply the units digit of 761 by 5. This is $1 \times 5 = 5$.
• Step 2: Multiply the tens digit of 761 by 5. This is $6 \times 5 = 30$. Write the 0 below and carry over the 3.
• Step 3: Multiply the hundreds digit of 761 by 5. This is $7 \times 5 = 35$. Add the carry-over 3 to get 38.

Write down the results from each multiplication:
From the units place, we have 5.
From the tens place, considering the carry, we have 0 (as $30$ is written as 0 with carry 3).
From the hundreds place, we have $38$.

Combining these, starting from the hundreds, tens, and units positions, gives us the final result of the multiplication: 3805.

Therefore, the product of 761 and 5 is $3805$.