When 335 is added to 5A7, the result is 8B2. 8B2 is divisible by 3. What is…

Answer: 4

Explanation:

• Units column: 5 + 7 = 12, so the units digit is 2 and there is a carry of 1 into the tens column.

• Tens column: 3 + A + 1 = 4 + A. This equals the tens digit B. If 4 + A ≥ 10 there would be a carry into the hundreds, but the hundreds column must sum to 8 without an extra carry (see next point). Therefore 4 + A ≤ 9, so A ≤ 5 and B = 4 + A.

• Hundreds column: 3 + 5 + (carry from tens). For the result to be 8, the carry from tens must be 0, confirming A ≤ 5.

• Divisibility by 3: The number 8B2 is divisible by 3, so 8 + B + 2 = 10 + B must be divisible by 3. That gives B ≡ 2 (mod 3), so B can be 2, 5, or 8.

• Combine B = 4 + A with the possible B values: 4 + A = 2 (impossible), 4 + A = 5 ⇒ A = 1, 4 + A = 8 ⇒ A = 4. Both A = 1 and A = 4 satisfy the constraints, and the largest possible A is 4.