If $p$ and $q$ represent digits, what is the maximum possible value of $q$ in the statement $$5p9 + 327 + 2q8 = 1114$$

### Concept & Logic This question requires maximizing a variable within a constrained sum. When the total sum of an equation is fixed, increasing one variable necessitates decreasing another. To maximize $q$, you must assign the absolute minimum possible value to $p$. ### Step-by-Step Solution * **Given Equation:** $5p9 + 327 + 2q8 = 1114$. * **Algebraic Breakdown:** Expand the numbers into their place values (hundreds, tens, and units) to isolate the unknown digits $p$ and $q$. $$(500 + 10p + 9) + 327 + (200 + 10q + 8) = 1114$$ * **Combine Constants:** $$10p + 10q + (500 + 327 + 200 + 9 + 8) = 1114$$ $$10p + 10q + 1044 = 1114$$ * **Simplify the Equation:** $$10(p + q) = 1114 - 1044$$ $$10(p + q) = 70$$ $$p + q = 7$$ * **Maximize q:** We know $p$ and $q$ are single digits ($0$ through $9$). To make $q$ as large as possible, we must make $p$ as small as possible. The lowest valid digit for $p$ is $0$. $$0 + q = 7 \implies q = 7$$ ### Exam Strategy & Shortcut **Column Addition Analysis:** Look only at the tens column to save time. Units sum: $9 + 7 + 8 = 24$ (write $4$, carry $2$). Hundreds sum: $5 + 3 + 2 = 10$. Since the final result is $1114$ (meaning the hundreds column is $11$), a carry of $1$ must have come from the tens place. Tens sum logic: $2 (\text{carry}) + p + 2 + q = 11$. This simplifies directly to $p + q = 7$. Set $p = 0$ to get max $q = 7$ in under 15 seconds. ### Common Pitfall Assuming that unknown digits in the middle of a number must be strictly greater than zero. The digit $0$ is perfectly valid for a middle place value (e.g., $509$). If you incorrectly assume $p$ must be at least $1$, you will get a maximum $q$ of $6$, which is a trap option. ### Final Answer Therefore, the correct answer is **7**.