What should be the maximum value of $Q$ in the following equation? $$5P9 - 7Q2 + 9R6 = 823$$

### Concept & Logic To find the maximum possible value of a subtracted variable ($Q$) in a fixed-sum equation, you must maximize the values of the added variables ($P$ and $R$). ### Step-by-Step Solution * **Given:** $5P9 - 7Q2 + 9R6 = 823$ * **Rearrange Terms:** Group the addition terms together for easier calculation. $$(5P9 + 9R6) - 7Q2 = 823$$ * **Place Value Expansion:** Break the terms into hundreds, tens, and units to isolate $P, Q,$ and $R$. $$(500 + 10P + 9) + (900 + 10R + 6) - (700 + 10Q + 2) = 823$$ * **Group Constants and Variables:** $$(500 + 900 - 700) + (9 + 6 - 2) + (10P + 10R - 10Q) = 823$$ $$700 + 13 + 10(P + R - Q) = 823$$ $$713 + 10(P + R - Q) = 823$$ * **Isolate Variables:** $$10(P + R - Q) = 823 - 713$$ $$10(P + R - Q) = 110$$ $$P + R - Q = 11$$ * **Maximize Q:** We know $P, Q,$ and $R$ are single digits ($0-9$). To make $Q$ as large as possible, we must make $P$ and $R$ as large as possible. Set $P = 9$ and $R = 9$. $$9 + 9 - Q = 11$$ $$18 - Q = 11$$ $$Q = 18 - 11 = 7$$ ### Exam Strategy & Shortcut **Tens Column Logic:** Ignore the hundreds and focus just on the tens and units carry-overs. Units: $9 - 2 + 6 = 13$. Write $3$ (matches the result $823$), carry over $+1$ to the tens column. Tens equation: $1 \text{ (carry)} + P - Q + R = \text{number ending in } 2$. Since $500 - 700 + 900 = 700$, and the final answer is $823$, the tens column must have generated a carry of $+1$ to the hundreds column. Therefore, the tens sum must literally equal $12$. $1 + P - Q + R = 12 \implies P + R - Q = 11$. Plug in max digits $P=9, R=9 \implies 18 - Q = 11 \implies Q = 7$. ### Common Pitfall Attempting to solve the equation vertically without converting the subtraction into a pure algebraic sum often leads to carry/borrow mistakes. Expanding into $10(P + R - Q)$ eliminates all ambiguity. ### Final Answer Therefore, the correct answer is **7**.