2025 CSAT — Q29

Write each numeral by place value. With $Q=3$:

The equation $PQR - PS = PPT$ becomes

Since $R,S,T$ are single digits, $R - S - T$ lies in $[-17, 7]$. So $20P-30$ must too: $P=1$ gives $-10$ (in range); $P=2$ gives $10$ (impossible, exceeds $7$). Hence $\boxed{P=1}$ and

Constraints: all of $P{=}1, Q{=}3, R, S, T$ distinct and non-zero; $T<6$ so $T \in \{2,4,5\}$ (excluding $1,3$); $R \ge 1$.

• $T=2$: $R = S - 8 \ge 1 \Rightarrow S = 9,\ R = 1 = P$ — rejected. (no valid pair)
• $T=4$: $R = S - 6 \ge 1 \Rightarrow S \in \{7,8,9\}$. $S=7\Rightarrow R=1=P$ ✗; $S=8\Rightarrow R=2$, digits $\{1,3,2,8,4\}$ distinct ✓; $S=9\Rightarrow R=3=Q$ ✗. → $(R,S)=(2,8)$.
• $T=5$: $R = S - 5 \ge 1 \Rightarrow S \in \{6,7,8,9\}$. $S=6\Rightarrow R=1=P$ ✗; $S=7\Rightarrow R=2$, $\{1,3,2,7,5\}$ ✓; $S=8\Rightarrow R=3=Q$ ✗; $S=9\Rightarrow R=4$, $\{1,3,4,9,5\}$ ✓. → $(R,S)=(2,7),(4,9)$.

Valid $(R,S)$: $(2,8),\ (2,7),\ (4,9)$ — three pairs. Verify each by subtraction: $132-18=114$ ✓, $132-17=115$ ✓, $134-19=115$ ✓.

Answer: (b) 3.

The decisive structural cut is the place-value identity reducing four unknowns to $R - S - T = 20P - 30$, which pins $P=1$ in one line (the only value keeping the right side inside the digit-range $[-17,7]$). After that the work is a tight, distinctness-checked enumeration over $T \in \{2,4,5\}$. The trap is forgetting that $R$ must differ from the already-used $P=1$ and $Q=3$ — three of the would-be solutions die exactly on that clash. CSAT rewards the candidate who carries the distinctness constraint through every branch rather than counting raw arithmetic solutions.