Vertical Addition: Addition of 3-digit and 1-digit numbers with the regrouping of both tens and ones

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

• Align 236 and 7 by their units (rightmost digits).
• Add the numbers starting from the rightmost digit (units), moving left:

Calculating step-by-step:
Step 1: Add the units digit. - Add $6$ (units place of 236) and $7$ (units place): $6 + 7 = 13$.
- This requires regrouping since the sum is more than $9$. Thus, write $3$ in the units place and carry over $1$ to the tens place.

Step 2: Add the tens digit. - Add the carry-over $1$ to the tens digit of 236, $3$: $1 + 3 = 4$.
- Place $4$ in the tens column.

Step 3: Add the hundreds digit. - There is only the hundreds digit $2$ from 236, and no carry-over: Thus result is also $2$ in the hundreds column.

Putting it all together, the sum is: $243$.

Therefore, the solution to the problem is $243$.

To solve this problem, we need to add the two numbers, $716$ and $8$, using vertical addition.

Step-by-step solution:

• Step 1: Add the units place digits: $6 + 8 = 14$. Here, $4$ is written in the units place, and we carry over $1$ to the tens place.

• Step 2: Add the tens place digits: $1 + 1 = 2$. As there was a carry from the previous step, the result is $2$ in the tens place.

• Step 3: Add the hundreds place digits: $7 + 0 = 7$. There is no carry to add, so the hundreds place remains $7$.

After combining these results, the number becomes $724$.

Therefore, the solution to the problem is $724$.

The correct answer choice is: Choice 4: $724$.

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

• Step 1: Add the ones place.
• Step 2: Handle any carrying to the tens place.
• Step 3: Add the tens place.
• Step 4: Handle any carrying to the hundreds place.
• Step 5: Add the hundreds place.

Now, let's work through each step:

Step 1: Consider the ones digits. For 872 and 9, the ones digits are 2 and 9.
Adding these gives us $2 + 9 = 11$.

Step 2: In base-10 addition, if the sum of digits in a place (in this case, the ones place) exceeds 9, we carry over to the next leftward digit. Thus, we write 1 in the ones place and carry over 1 to the tens place.

Step 3: Now, we add the tens digits. For the tens, we have 7 (from 872) and the 1 from our carry over.
Thus, $7 + 1 = 8$.

Step 4: After adding the tens digits, check if there's any carry over. Since 8 is not greater than 9, no further carrying is needed.

Step 5: Finally, add the hundreds place digits. For the hundreds, we have 8 (from 872), and there was no carry from the tens, giving $8 + 0 = 8$.

Combining these sum results, we have the final sum as: Hundreds: 8
Tens: 8
Ones: 1
Therefore, the sum of 872 and 9 is $881$.

Therefore, the solution to the problem is $881$.

The choice that corresponds to this solution is Choice 4: $881$.

To solve this problem, we'll perform the following steps in vertical addition:

1. Write the numbers vertically, aligning them by their place values:
$\begin{aligned} &543 \\ + &~~~~8 \\ &\underline{\phantom{776}} \end{aligned}$

2. Begin with the units place (rightmost column):
Add $3$ (from $543$) and $8$, which equals $11$. Since this number is greater than $9$, we have to regroup. Write down $1$ and carry over $1$ to the tens place.

3. Move to the tens place:
Add $4$ (from $543$) plus the carry-over $1$, which equals $5$. Write $5$ under the tens place column.

4. Finish with the hundreds place:
Add $5$ (from $543$), and since there is no carry-over, write $5$ under the hundreds place column.

Therefore, after completing the addition process, we have the result:

$\begin{aligned} &543 \\ + &~~~~8 \\ &\underline{551} \end{aligned}$

The solution to the addition problem is $551$.

To solve the problem of adding 917 and 5, follow these step-by-step instructions:

• Step 1: Line up the numbers vertically by their place values. The hundreds place, tens place, and units place should be aligned:

$\begin{aligned} &917 \\ +& \\ &~~~5 \\ &\underline{\phantom{776}} \\ \end{aligned}$

• Step 2: Begin with the units place (rightmost column). Add $7$ and $5$. The sum is $12$.

Since $12$ is greater than $9$, we write down $2$ and carry over $1$ to the tens place.

• Step 3: Proceed to the tens place. Add the digits $1$ (the carried over value) and $1$ (from 917). This gives us $2$.

We write $2$ in the tens place.

• Step 4: Finally, add the hundreds place. Add $9$ (from 917). As there is no shipment here due to the tens column sum, write the $9$ directly.

After completing all the places, the final result is:

$\begin{aligned} &917 \\ +& \\ &~~~5 \\ &\underline{922} \\ \end{aligned}$

Therefore, the answer to the problem is $\boxed{922}$.

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

• Step 1: Align the numbers 173 and 18 by their rightmost digits. Write 173 above 18.
• Step 2: Start adding from the ones column (rightmost). 3 (ones of 173) + 8 (ones of 18) = 11.
• Step 3: Write down the 1 in the ones place and carry over the remaining 1 to the tens column.
• Step 4: Move to the tens column. Add 7 (tens of 173) + 1 (tens of 18) = 8, plus the carried over 1 = 9.
• Step 5: Write 9 in the tens place.
• Step 6: Move to the hundreds column. Add 1 (hundreds of 173) with the carry, which is zero here, so it remains 1.

Therefore, the final sum of 173 and 18 is $191$.

To solve this vertical addition problem, follow these steps:

• Step 1: Align the numbers according to place value: hundreds, tens, and ones.
  The numbers are:
  276
  + 16

• Step 2: Start adding from the rightmost column (the ones place):
  - Add 6 (from 276) and 6 (from 16) = 12.
  - Write down 2 in the ones place and carry over 1 to the tens place.

• Step 3: Move to the tens column:
  - Add 7 (from 276), 1 (carried), and 1 (from 16) = 9.
  - Write down 9 in the tens place.

• Step 4: Finally, move to the hundreds column:
  - Since there's no carried number here, add 2 (from 276) as it is.
  - Write down 2 in the hundreds place.

After performing these operations, the final sum is written as:

292

Therefore, the correct answer to the problem is $292$.

To solve this problem, we'll carry out vertical addition:

• Step 1: Add the rightmost digits (units place): $7 + 5 = 12$. Since 12 is more than 9, we write 2 in the units place and carry over 1 to the tens place.
• Step 2: Add the tens place along with the carried-over digit: $1 + 1 + 4 = 6$. Write 6 in the tens place.
• Step 3: Add the hundreds place: $3 + (0 \text{ since there is no hundreds digit from 45}) = 3$. Write 3 in the hundreds place.

Therefore, the sum of the numbers 317 and 45 is $\mathbf{362}$.

The correct choice from the given options that matches this sum is choice 4: $362$.

To solve this problem, we need to add the numbers 728 and 56 using vertical addition. Let's break this down step by step:

• Step 1: Add the ones place digits. Here, we add $8 + 6 = 14$. The digit 4 remains in the ones place, and we carry over the 1 to the tens place.
• Step 2: Add the tens place. Now, add the tens digits, taking into account the carry-over: $2 + 5 + 1 = 8$. Write 8 in the tens place.
• Step 3: Add the hundreds place. Since there are no more digits to add to the hundreds place and no carry over, we keep the hundreds digit as is: $7$.

Thus, the sum of 728 and 56 is $784$.

Therefore, the correct answer is choice 2: 784.