UPSC Prelims 2025 CSAT, question 7

Solution.

Answer: (c) Three

1. Assume without loss of generality that $1 < p \le q \le r$; since $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1 \le \frac{3}{p}$, we must have $p \in \{2, 3\}$.
2. For $p = 3$, the only solution with $3 \le q \le r$ is $(3, 3, 3)$, giving $p + q + r = 9$.
3. For $p = 2$, we need $\frac{1}{q} + \frac{1}{r} = \frac{1}{2}$, which gives $(q - 2)(r - 2) = 4$; the solutions are $(2, 4, 4)$ with sum $10$ and $(2, 3, 6)$ with sum $11$, yielding $3$ possible values ($9, 10,$ and $11$).