In the equation formed with digit placeholders 8A9 – 6B2 + 4C6 = 723, A, B, and C represent single digits (0 to 9). What is the maximum possible value of the digit B for which this equation holds true?

Introduction / Context:
This question involves working with numbers written using digit placeholders and solving a simple equation in digits. The three digit numbers 8A9, 6B2, and 4C6 are formed by inserting digits A, B, and C in the tens place. We know the overall equation 8A9 – 6B2 + 4C6 = 723 is satisfied, and our goal is to determine the largest possible value of the digit B for which this holds. This tests understanding of place value and basic algebraic manipulation.

Given Data / Assumptions:

• 8A9, 6B2, and 4C6 are three digit numbers.
• 8A9 = 800 + 10A + 9.
• 6B2 = 600 + 10B + 2.
• 4C6 = 400 + 10C + 6.
• The equation is (800 + 10A + 9) – (600 + 10B + 2) + (400 + 10C + 6) = 723.
• A, B, and C are digits from 0 to 9.

Concept / Approach:
We first expand each three digit number in terms of hundreds, tens, and units, and then simplify the entire expression. By collecting the constant parts and the terms involving A, B, and C, we reduce the equation to a simple linear relation between A, B, and C. Then we use the fact that A, B, and C are digits to find the maximum possible B that satisfies this relation for some valid choices of A and C.

Step-by-Step Solution:
Step 1: Write each number in expanded form.Step 2: 8A9 = 800 + 10A + 9.Step 3: 6B2 = 600 + 10B + 2.Step 4: 4C6 = 400 + 10C + 6.Step 5: Substitute these into the equation: (800 + 10A + 9) - (600 + 10B + 2) + (400 + 10C + 6) = 723.Step 6: Combine constant terms: 800 - 600 + 400 = 600 for the hundreds and 9 - 2 + 6 = 13 for the units, so the constant part is 600 + 13 = 613.Step 7: Combine the tens terms: 10A - 10B + 10C = 10(A - B + C).Step 8: The equation becomes 613 + 10(A - B + C) = 723.Step 9: Subtract 613 from both sides: 10(A - B + C) = 723 - 613 = 110.Step 10: Divide by 10: A - B + C = 11.Step 11: Rearrange to express B in terms of A and C: B = A + C - 11.Step 12: Since A and C are digits from 0 to 9, the maximum possible sum A + C is 9 + 9 = 18.Step 13: Therefore, maximum B = 18 - 11 = 7.Step 14: Check that B = 7 is valid, for example by taking A = 9 and C = 9, which gives B = 9 + 9 - 11 = 7 (a valid digit).

Verification / Alternative check:
Using A = 9, B = 7, and C = 9, compute each three digit number explicitly. 8A9 = 8 * 100 + 9 * 10 + 9 = 800 + 90 + 9 = 899. 6B2 = 600 + 7 * 10 + 2 = 672. 4C6 = 400 + 9 * 10 + 6 = 496. Now evaluate 899 - 672 + 496. First 899 - 672 = 227, then 227 + 496 = 723, which matches the given right side exactly. This confirms that B = 7 is attainable and that it is the maximum possible value consistent with the digit constraints.

Why Other Options Are Wrong:
Option 4: This would correspond to a smaller sum A + C and is not the maximum possible value of B.Option 1: Too small and does not exploit the maximum range of A and C.Option 5: Although attainable for some A and C, it is not the largest value allowed by the condition A - B + C = 11.Option 6: Also below the maximum possible B of 7.

Common Pitfalls:
Students may forget that A, B, and C are digits restricted to 0 through 9 and might try values outside this range. Another error is miscalculating the combined constant term or mishandling the tens coefficients. Carefully expanding each three digit number and collecting like terms prevents such mistakes. Finally, some may assume that A, B, and C must be distinct, which is not stated in the problem; we only need them to be digits.

Final Answer:
The maximum possible value of B is 7.