How many possible values of (p + q + r) are there satisfying 1/p + 1/q + 1/r = 1, where p, q and r are natural numbers (not necessarily distinct)?
Answer & explanation
Answer: (c) Three
Only three unordered triples satisfy 1/p + 1/q + 1/r = 1: (3, 3, 3), (2, 4, 4) and (2, 3, 6). Their sums 9, 10 and 11 are all different, so p + q + r can take three values.
- Arrange so that p ≤ q ≤ r. Then 1/p is the largest of the three fractions, so 1/p ≥ 1/3, i.e. p ≤ 3; and p ≥ 2 because 1/p must be less than 1.
- p = 3: 1/q + 1/r = 2/3 with q, r ≥ 3 forces q = r = 3. Sum = 9.
- p = 2: 1/q + 1/r = 1/2 with q ≤ r gives (q, r) = (3, 6) or (4, 4). Sums = 11 and 10.
- Possible values of p + q + r: 9, 10 and 11 — three values.
- Check: 1/2 + 1/3 + 1/6 = 1, 1/2 + 1/4 + 1/4 = 1, 1/3 + 1/3 + 1/3 = 1.
Remember · For unit-fraction equations, sort the variables and bound the smallest one first; the cases then collapse quickly.
Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·