Q19. 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)?

Given the equation: PQR - PS = PPT This can be written in expanded form: (100P + 10Q + R) - (10P + S) = (100P + 10P + T) 100P + 10Q + R - 10P - S = 110P + T 90P + 10Q + R - S = 110P + T Substitute Q = 3: 90P + 10(3) + R - S = 110P + T 90P + 30 + R - S = 110P + T Rearrange the terms to isolate R - S: R - S = 110P - 90P + T - 30 R - S = 20P + T - 30 We are given that P, Q, R, S, T are distinct non-zero digits. Since Q = 3, none of P, R, S, T can be 3.

Also, T < 6, so T can be 1, 2, 4, 5 (T cannot be 3). The range for R - S (where R and S are distinct non-zero digits, not 3) is from (1-9) = -8 to (9-1) = 8. So, -8 ≤ R - S ≤ 8. Let's test possible values for P (P is a non-zero digit, not 3): Case 1: P = 1 R - S = 20(1) + T - 30 = T - 10 - If T = 1: R - S = 1 - 10 = -9.

This is outside the range [-8, 8]. No solution. - If T = 2: R - S = 2 - 10 = -8. Possible (R, S) pair: (1, 9). But R=1, P=1, and digits must be distinct. So, (1, 9) is not valid. - If T = 4: R - S = 4 - 10 = -6.

Possible (R, S) pairs: (1, 7), (2, 8), (3, 9). - (1, 7): R=1, P=1 (not distinct). Invalid. - (2, 8): R=2, S=8. Digits (P=1, Q=3, R=2, S=8, T=4) are distinct and non-zero. Valid! -> (R,S) = (2,8) is a solution.

- (3, 9): R=3, Q=3 (not distinct). Invalid. - If T = 5: R - S = 5 - 10 = -5. Possible (R, S) pairs: (1, 6), (2, 7), (3, 8), (4, 9). - (1, 6): R=1, P=1 (not distinct). Invalid. - (2, 7): R=2, S=7.

Digits (P=1, Q=3, R=2, S=7, T=5) are distinct and non-zero. Valid! -> (R,S) = (2,7) is a solution. - (3, 8): R=3, Q=3 (not distinct). Invalid. - (4, 9): R=4, S=9. Digits (P=1, Q=3, R=4, S=9, T=5) are distinct and non-zero.

Valid! -> (R,S) = (4,9) is a solution. Case 2: P = 2 R - S = 20(2) + T - 30 = 40 + T - 30 = T + 10. Since T can be at most 5, T + 10 will be at least 1 + 10 = 11. This value (11) is outside the range [-8, 8] for R - S.

So, no solutions for P = 2. For any P ≥ 4 (since P cannot be 3), 20P will be even larger, making R-S even larger than 8. So, no solutions for P ≥ 4. Thus, the only possible values for (R, S) are (2, 8), (2, 7), and (4, 9).

There are 3 possible values for (R, S). This is a challenging cryptarithmetic problem that requires careful algebraic manipulation, systematic case analysis, and attention to constraints (distinct non-zero digits), typical of higher-difficulty logical reasoning questions in UPSC CSAT.