Consider the following multiplication problem: (PQ) × 3 = RQQ , where P, Q and R are different digits and R ≠ 0 . What is the value of (P + R) ÷ Q ?

The problem is (PQ) x 3 = RQQ. This can be written as (10P + Q) x 3 = 100R + 10Q + Q. 1. Analyze the units digit: For the units digit of RQQ to be Q, the units digit of Q x 3 must be Q. This occurs only for Q = 0 (0 x 3 = 0) or Q = 5 (5 x 3 = 15, units digit is 5). 2. Eliminate Q = 0: If Q = 0, the equation becomes (P0) x 3 = R00, which simplifies to 30P = 100R, or 3P = 10R. Since P and R are single non-zero digits (P cannot be 0 as PQ is a two-digit number, and R is given as non-zero), there is no integer solution for P and R. Thus, Q cannot be 0. 3. Determine Q = 5: Therefore, Q must be 5. When Q = 5, 5 x 3 = 15. This means the units digit of RQQ is 5 (which is Q), and there is a carry-over of 1 to the tens place. 4. Analyze the tens digit: For the tens digit of RQQ to be Q (which is 5), the result of (P x 3) + 1 (carry-over) must have a units digit of 5. This implies that P x 3 must end in 4. The only single digit P (from 1 to 9) for which P x 3 ends in 4 is P = 8 (since 8 x 3 = 24). 5. Determine R: With P = 8 and Q = 5, the multiplication is (85) x 3. 85 x 3 = 255. Comparing this to RQQ, we get R = 2. 6. Verify conditions: P = 8, Q = 5, R = 2. These are different digits, and R is not 0. All conditions are met. 7. Calculate the final expression: We need to find the value of (P + R) / Q. (8 + 2) / 5 = 10 / 5 = 2. The value is 2. The final answer is B