Given:

• PQR − PS = PPT
• Q = 3; digits P, Q, R, S, T are distinct and non-zero; T < 6.

Translate to digits: PQR = $100P + 30 + R$ PS = $10P + S$ PPT = $110P + T$

Equation: $(100P + 30 + R) − (10P + S) = 110P + T$ $⇒ 90P + 30 + R − S = 110P + T$ $⇒ R − S = 20P + T − 30$ (★)

Range bound: Since R, S ∈ {1,…,9} and R ≠ S, we must have R − S ∈ {−8, −7, …, −1, 1, …, 8}.

Also T < 6 and T ≠ 3 (distinct from Q), and T ≠ P (distinctness).

Check P values:

• If P ≥ 2, RHS = $20P + T − 30 ≥ 40 + 1 − 30 = 11 > 8$ ⇒ impossible.
• Hence P = 1.

Then (★) becomes: R − S = $(20 \cdot 1 + T − 30) = T − 10$. With T < 6 and T ≠ 1 (distinct from P) and T ≠ 3, possible T ∈ {2, 4, 5}.

For each T, solve R − S = T − 10 with digit/distinctness constraints (R,S ≠ 1,3,T; R,S ∈ {1…9}, R ≠ S):

1. T = 2 ⇒ R − S = −8 ⇒ R = S − 8. Possible S is 9 ⇒ R = 1, but R = 1 is disallowed (P = 1). ⇒ No solution.

2. T = 4 ⇒ R − S = −6 ⇒ R = S − 6. S ∈ {7,8,9}:

   • S=7 → R=1 (disallowed)
   • S=8 → R=2 (allowed) → (R,S) = (2,8)
   • S=9 → R=3 (disallowed, Q=3)

3. T = 5 ⇒ R − S = −5 ⇒ R = S − 5. S ∈ {6,7,8,9}:

   • S=6 → R=1 (disallowed)
   • S=7 → R=2 (allowed) → (2,7)
   • S=8 → R=3 (disallowed)
   • S=9 → R=4 (allowed) → (4,9)

Valid (R,S): (2,8), (2,7), (4,9) → 3 pairs.

Therefore, the number of possible values of (R, S) is 3.