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)?

1. A. 2
2. B. 3
3. C. 4
4. D. More than 4

Answer: B

Explanation

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 (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.