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 is a 3-digit number with hundreds digit P, tens digit Q=3, units digit R. PS is 2-digit with tens P, units S. PPT is 3-digit with hundreds and tens both P, units T. PQR − PS = PPT means 100P + 30 + R − (10P + S) = 110P + T, so 30 + R − S = 20P + T, giving R − S = 20P + T − 30. For P=1: R − S = T − 10 (negative since T<6), impossible as R, S are non-zero digits with R−S ≥ −8. For P=2: R − S = T + 10, impossible since max R−S = 8. So only P=1 works, giving S − R = 10 − T. With T ∈ {2, 4, 5} (T<6, distinct from P=1, Q=3, non-zero), checking distinct digits constraint yields 3 valid (R, S) pairs.