The sum of three consecutive integers is equal to their product. How many such possibilities are…

Answer: (c) Only three

Writing the integers as n − 1, n, n + 1 gives 3n = n(n² − 1), so n = 0, 2 or −2. That produces three triples: (−1, 0, 1), (1, 2, 3) and (−3, −2, −1).

1. Let the integers be n − 1, n, n + 1. Sum = 3n; product = n(n² − 1).
2. 3n = n(n² − 1) gives n = 0 or n² − 1 = 3, i.e. n = 0, 2 or −2.
3. The triples are (−1, 0, 1), (1, 2, 3) and (−3, −2, −1), with sums 0, 6, −6 equal to their products.
4. So there are exactly three possibilities. (Testing only positive integers finds just 1, 2, 3.)

Remember · ‘Integers’ includes zero and negatives; solve the algebra instead of testing only positive numbers.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). · Report an error