\( 28,447 + 6 \square , 54 \square = 95,9 \square 0 \)

To solve this puzzle, we need to fill in the squares with digits that make both sides of the equation equal. The two parts we need to solve for are the unknown digits in \( 6 \square \) and \( 54 \square \). Let's break it down: starting from the rightmost digit, we know \( 7 + \square = \square + 0 \). This means the last digit calculation should result in a number that ends in 0 - thus the last square must be 3. Now looking at the tens place, we have \( 4 + 5 + 1 (carry) = 9 \). The tens place of the unknown sum should also remain intact. For the hundreds place with \( 8 + 6 + 1 (carry from the tens) \), we get \( 15 \), which implies a carry of 1 again. Repeating this for thousands (where \( 2 + 0 + 1 (carry) = 9 \) leads us to find the missing digit in \( 6 \square \) as \( 6 \)), the completed numbers would be: \( 28,447 + 6 3 = 95,900 \). So \( 28,447 + 63 + 54 \). Thus, after breaking it down, we find the numbers to fill in are 6 and 3!