The sum of three consecutive integers is equal to their product. How many such possibilities are there?
- (a) Only one
- (b) Only two
- (c) Only three ✓ UPSC's answer
- (d) No such possibility is there
Why the answer is (c)
• Let the three consecutive integers be n-1, n, and n+1.
• Their sum is 3n, and their product is n(n-1)(n+1) = n(n^2 - 1).
• Equating sum and product gives 3n = n(n^2 - 1), so n(n^2 - 4) = 0.
• Therefore n = 0, n = 2, or n = -2.
• These give the triples (-1, 0, 1), (1, 2, 3), and (-3, -2, -1), so there are exactly three possibilities, matching option (c).
Why the other options are wrong
- (a) Only one
- Only one is incorrect because the equation gives three valid middle values: 0, 2, and -2.
- (b) Only two
- Only two is incorrect because all three solutions n = 0, n = 2, and n = -2 satisfy the condition.
- (d) No such possibility is there
- No such possibility is incorrect because (1, 2, 3), (-1, 0, 1), and (-3, -2, -1) are valid examples.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2022, held on 5 June 2022. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.