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…
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 $(R, S)$?
2
3
4
More than 4
Solution
PQR is the 3-digit number $\overline{P3R}$, PPT is $\overline{PPT}$, and PS is $\overline{PS}$. Solving $\overline{P3R} - \overline{PS} = \overline{PPT}$ with the digit-place borrow conditions, and applying the constraints $Q=3$, $T<6$, all digits distinct and non-zero, yields exactly 3 valid $(R,S)$ pairs.