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)?

To solve the equation PQR minus PS equals PPT, we can write it in terms of place values:

100P + 10Q + R minus (10P + S) = 100P + 10P + T

By simplifying this equation, we get: 10Q + R minus S = 10P + T

Given Q = 3, the equation becomes: 30 + R minus S = 10P + T

Since P, Q, R, S, and T are distinct non-zero digits, and T is less than 6 (1, 2, 4, or 5, as 3 is taken by Q), we test possible values for P:

If P = 2: 30 + R minus S = 20 + T 10 + R minus S = T R + 10 = S + T

We check pairs for (R, S) ensuring all digits are distinct and non-zero (excluding 2 and 3):

1. If T = 1: R + 10 = S + 1. No solution for single digits.
2. If T = 4: R + 10 = S + 4 -> S minus R = 6. Pairs: (1,7), (2,8), (3,9). Only (1,7) works as 2 and 3 are used.
3. If T = 5: R + 10 = S + 5 -> S minus R = 5. Pairs: (1,6), (4,9). Both work.

If P = 3: Not possible as P and Q must be distinct.

If P = 1: 30 + R minus S = 10 + T 20 + R minus S = T This is impossible as T would be greater than 20.

If P = 4 or higher: 30 + R minus S = 40 + T R minus S = 10 + T This is impossible for single digits.

The valid pairs for (R, S) are (1,7), (1,6), and (4,9). Thus, there are 3 possible values.

Correct Answer: B