What would be the maximum value of $Q$ in the following equation? $$5P7 + 8Q9 + R32 = 1928$$

### Concept & Logic To find the maximum possible value of a digit in an addition problem, you must evaluate the maximum possible carry-overs from previous columns. Do not assume the carry-over is limited to $1$; verify if larger carry-overs are mathematically possible. ### Step-by-Step Solution * **Given:** $5P7 + 8Q9 + R32 = 1928$. * **Units Column:** Add the known unit digits. $7 + 9 + 2 = 18$. Write down $8$ and carry over $1$ to the tens column. * **Tens Column:** Form the sum using the carry. $1 \text{ (carry)} + P + Q + 3 = P + Q + 4$ Since the tens digit in the final answer is $2$, the sum $P + Q + 4$ must end in $2$. It could be $12$ (generating a carry of $1$) or $22$ (generating a carry of $2$). * **Case 1 (Sum is 12):** $P + Q + 4 = 12 \implies P + Q = 8$. (Carry to hundreds is $1$). * **Case 2 (Sum is 22):** $P + Q + 4 = 22 \implies P + Q = 18$. (Carry to hundreds is $2$). This is possible if $P=9$ and $Q=9$. * **Hundreds Column Verification:** Let us check if Case 2 is valid for the hundreds column. If carry is $2$, the hundreds equation is: $2 + 5 + 8 + R = 19$. $15 + R = 19 \implies R = 4$. * **Conclusion:** Since $R = 4$ is a valid single digit, the scenario where $P = 9$ and $Q = 9$ is entirely valid ($597 + 899 + 432 = 1928$). Thus, the maximum possible value for $Q$ is $9$. ### Exam Strategy & Shortcut **Theoretical Maximum Test:** The absolute maximum value for any single digit is $9$. Instead of building equations from the ground up, assume $Q = 9$ immediately and try to force the equation to work. If $Q = 9$, we need a high value for $P$ to make the sum valid. Try $P = 9$. Tens sum becomes $1 (\text{carry}) + 9 + 9 + 3 = 22$. We write $2$ (matches the result) and carry $2$. Hundreds becomes $2 + 5 + 8 + R = 19 \implies R = 4$. Since everything resolves to single, valid digits, $Q = 9$ is proven correct instantly. ### Common Pitfall Many students naturally assume the tens column sum $P + Q + 4$ must be $12$ because they are accustomed to carry-overs of $1$. This leads them to $P + Q = 8$, concluding the max of $Q$ is $8$ (Option B). Always check if a carry of $2$ is possible when maximizing variables! ### Final Answer Therefore, the correct answer is **9**.