Let PQR be a 3-digit number, PPT be a 3-digit number and PS be a 2-digit number, where P, Q, R, S, T are distinct non-zero digits. Further, PQR − PS = PPT. If Q = 3 and T < 6, then what is the number of possible values of (R, S) ?

PQR = 100P + 10Q + R = 100P + 30 + R. PS = 10P + S. PPT = 110P + T. Equation: (100P + 30 + R) − (10P + S) = 110P + T, so 90P + 30 + R − S = 110P + T, giving R − S − T = 20P − 30. For P = 1: R − S − T = −10, so S + T − R = 10. With T < 6, T ∈ {1,2,3,4,5} (and T ≠ Q=3, T≠P=1), so T ∈ {2,4,5}. For each: T=2: S−R = 8, so (S,R) = (9,1)—but R≠P=1, invalid; (S,R)=(8,0) but R≠0. T=4: S−R = 6: (S,R)=(9,3) R≠Q=3 invalid; (8,2)✓; (7,1) R≠1 invalid; (6,0) invalid. T=5: S−R = 5: (S,R)=(9,4)✓; (8,3) invalid; (7,2)✓; (6,1) invalid. Valid (R,S) pairs: (2,8), (4,9), (2,7) — that gives 3 possible pairs. Hence the answer is 3.