2025 CSAT — Q7

Order them $p \le q \le r$ and bound the smallest denominator. Since the three terms sum to $1$, the largest term $\tfrac1p$ is at least $\tfrac13$, so $p \le 3$; and $\tfrac1p < 1$ forces $p \ge 2$. Hence $p \in \{2,3\}$.

Case $p = 3$: then $\tfrac1q+\tfrac1r = \tfrac23$ with $q \ge 3$. Now $\tfrac1q \ge \tfrac13$ forces $q = 3$, giving $\tfrac1r = \tfrac13$, so $r = 3$. Solution $(3,3,3)$, sum $= 9$.

Case $p = 2$: then $\tfrac1q+\tfrac1r = \tfrac12$ with $q \ge 2$. Bounding $\tfrac1q \ge \tfrac14$ gives $q \in \{3,4\}$:

• $q = 3 \Rightarrow \tfrac1r = \tfrac16 \Rightarrow r = 6$: solution $(2,3,6)$, sum $= 11$.
• $q = 4 \Rightarrow \tfrac1r = \tfrac14 \Rightarrow r = 4$: solution $(2,4,4)$, sum $= 10$.

The three unordered solutions are $(3,3,3),\ (2,4,4),\ (2,3,6)$, giving sums $\{9, 10, 11\}$ — three distinct values of $p+q+r$.

Answer: (c) Three.

The engine of every unit-fraction equation is the bounding move: with $p \le q \le r$, the smallest denominator is squeezed into a tiny range ($p \le k/(\text{RHS})$). Here it pins $p \in \{2,3\}$ in one line and the rest cascades. The planted trap is the ordered-vs-unordered confusion in option (d): the triples number more than three, but the question asks how many distinct values of the sum — and that is exactly $|\{9,10,11\}| = 3$. Read the target quantity precisely before counting.