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Let's break down the addition puzzle to determine which digits replace \( A \), \( B \), and \( C \).

We have the following setup:

\[
\begin{array}{c@{}c@{}c@{}c@{}c}
  2 & 1 & . & 3 & 6 \\
+ 3 & 3 & . & 9 & A \\
\hline
  5 & 5 & . & B & 3 \\
\end{array}
\]

### Step 1: Focus on the Units Column (rightmost digits)

From the units column, we have:
\[
6 + A = 3 \ (\text{or possibly } 13 \text{ with carryover}).
\]
Since \( 6 + A \) cannot directly give 3, there must be a carry of 1 from the tenths column. So:
\[
6 + A + 1 = 13,
\]
which means:
\[
A = 6.
\]

### Step 2: Focus on the Tenths Column (next to last digits)

Now, in the tenths column, we have:
\[
3 + 9 + C = B \ (\text{plus a possible carry of 1 from the units column}).
\]
With the carry of 1 from the units column, we get:
\[
3 + 9 + 1 = 13.
\]
Thus, \( B = 3 \), and there is another carryover of 1 to the hundredths column.

### Step 3: Focus on the Hundredths Column (third column from right)

Here we have:
\[
1 + 1 + C = 5.
\]
This means:
\[
C = 3.
\]

### Conclusion:

The digits that replace \( A \), \( B \), and \( C \) are:

- \( A = 6 \)
- \( B = 3 \)
- \( C = 3 \)

Would you like further details or have any questions?

Here are some related questions to consider:

1. How do you handle addition with multiple carryovers?
2. Can there be a solution where no carryovers are involved?
3. How can this method of addition be applied to larger numbers?
4. What happens if the sum exceeds the number of digits in one of the numbers?
5. What strategies can help avoid mistakes when dealing with carryovers?

**Tip**: When adding multiple columns with carryovers, always start from the rightmost digit and work your way left to avoid confusion.

Which digits replace A, B, and C in the boxes?

Learn how to solve a 3-column addition puzzle where A, B, and C represent unknown digits. This step-by-step guide demonstrates the use of carryover and basic addition principles to solve for the unknowns, suitable for students in Grades 4-6.

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