Vertical Addition: Addition of 3-digit and 2-digit numbers with regrouping beyond 1000

To solve this problem, we will perform vertical addition as follows:

• First, align the numbers:
          569
  +      437
         ----

• Add the units' digit (rightmost column):
     9 (from 569) + 7 (from 437) = 16
  Write the 6 in the units place of the result and carry over the 1.

• Add the tens' digit:
     6 (from 569) + 3 (from 437) + 1 (carry) = 10
  Write the 0 and carry over the 1.

• Add the hundreds' digit:
     5 (from 569) + 4 (from 437) + 1 (carry) = 10
  Write the entire number since this is the leftmost position.

Hence, the final sum is:
       1006

Therefore, the solution to the problem is $1006$.

To solve this problem, we'll engage in vertical addition for the given numbers $743$ and $259$:

• Step 1: Align the numbers by their place value (units, tens, and hundreds).
• Step 2: Add each column starting from the right, taking care to carry over if needed.

Let's perform the addition:
Starting from the rightmost column (units):

• Add $3$ (from $743$) and $9$ (from $259$): $3 + 9 = 12$.
  Place $2$ in the units place and carry over $1$ to the tens column.
• Move to the tens column:
  Add $4$ (from $743$), $5$ (from $259$), and the carryover $1$: $4 + 5 + 1 = 10$.
  Place $0$ in the tens place and carry over $1$ to the hundreds column.
• Finally, add the hundreds column:
  Add $7$ (from $743$), $2$ (from $259$), and the carryover $1$: $7 + 2 + 1 = 10$.
  Place $0$ in the hundreds place and $1$ in the thousands place.

The final result of adding $743 + 259$ is $1002$.

Comparing this with the answer choices, the correct choice is (1): $1002$.

Therefore, the solution to the problem is $1002$.

To solve this problem, we will perform vertical addition starting from the rightmost digit:

• Step 1: Add the units digits.
  $9 + 7 = 16$. Write down $6$ and carry over $1$ to the tens place.
• Step 2: Add the tens digits.
  $8 + 1 = 9$, now add the carry $1$, so $9 + 1 = 10$. Write down $0$ and carry over $1$ to the hundreds place.
• Step 3: Add the hundreds digits.
  $6 + 3 = 9$, now add the carry $1$, so $9 + 1 = 10$. Write down $0$ in the hundreds place and write down $1$ in the thousands place (since it's the final column, the carry becomes a leading digit).

Therefore, the sum of $689$ and $317$ is $\mathbf{1006}$.

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

• Step 1: Add the units digits of both numbers.
• Step 2: Handle any regrouping if necessary and add the tens digits.
• Step 3: Handle any regrouping if necessary and add the hundreds digits.

Now, let's work through each step:

Step 1: Add the units digits of both numbers:
$8 + 9 = 17$. Here, 7 is placed in the units place, and 1 is carried over to the tens column.

Step 2: Add the tens digits along with the carry-over:
$4 + 5 + 1 = 10$. Here, 0 is placed in the tens place, and 1 is carried over to the hundreds column.

Step 3: Add the hundreds digits along with the carry-over:
$5 + 4 + 1 = 10$. Here, 0 is placed in the hundreds place, and 1 is carried over to the thousands column.

The final sum is $1007$, where 1 is now the digit in the thousands place.

Therefore, the solution to the problem is $1007$.

To solve this problem of adding $826$ and $178$ using vertical addition, we proceed as follows:

• First, we align the numbers vertically, ensuring that digits are in their respective places:

$\begin{array}{c} & 8 & 2 & 6 \\ + & 1 & 7 & 8 \\ \hline \end{array}$

• Step 1: Add the units (rightmost) column: $6 + 8 = 14$. Place the $4$ in the units column of the result and carry over the $1$ to the tens column.

• Step 2: Add the tens column: $2 + 7 = 9$ and add the carried over $1$: $9 + 1 = 10$. Place $0$ in the tens column of the result and carry over the $1$ to the hundreds column.

• Step 3: Add the hundreds column: $8 + 1 = 9$ and add the carried over $1$: $9 + 1 = 10$. Write down the entire sum, $10$, in the hundreds and beyond place.

The resulting sum is $1004$.

Thus, the solution to the problem is $1004$.