तीन क्रमागत पूर्णांकों का योग उनके गुणनफल के बराबर हो, ऐसी कितनी संभावनाएँ हो सकती हैं?

20223Mprelimsupscyear-2022p2quantitative-aptitudequantitative-abilityprobability
A.केवल एक
B.केवल दो
C.केवल तीनसही
D.ऐसी कोई संभावना नहीं है
व्याख्या एवं हल

सही उत्तर: (C) केवल तीन

माना तीन क्रमागत पूर्णांक n−1,n,n+1n - 1, n, n + 1 हैं। इनका योग 3n3n तथा गुणनफल n(n2−1)n(n^2 - 1) है। दोनों को बराबर रखने पर: 3n=n(n2−1)  ⟹  n(n2−4)=03n = n(n^2 - 1)   ⟹  \implies n(n^2 - 4) = 0, जिससे n=0,2,−2n = 0, 2, -2 प्राप्त होता है। इसके अनुसार तीन पूर्णांक समुच्चय बनते हैं: (1) {−1,0,1}\{-1, 0, 1\} (योग 0, गुणनफल 0), (2) {1,2,3}\{1, 2, 3\} (योग 6, गुणनफल 6), तथा (3) {−3,−2,−1}\{-3, -2, -1\} (योग -6, गुणनफल -6)। अतः ऐसी केवल तीन संभावनाएँ हैं।

In English (Question & Model Answer)

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 threeCorrect
D.No such possibility is there

Correct Answer: (C) Only three

Let the three consecutive integers be n−1,n,n+1n - 1, n, n + 1. Their sum is 3n3n and their product is (n−1)n(n+1)=n(n2−1)(n - 1)n(n + 1) = n(n^2 - 1). Setting them equal gives 3n=n(n2−1)  ⟹  n(n2−4)=03n = n(n^2 - 1)   ⟹  \implies n(n^2 - 4) = 0, yielding solutions n=0n = 0, n=2n = 2, and n=−2n = -2. The corresponding sets of integers are: (1) {−1,0,1}\{-1, 0, 1\} with sum 0 and product 0; (2) {1,2,3}\{1, 2, 3\} with sum 6 and product 6; and (3) {−3,−2,−1}\{-3, -2, -1\} with sum -6 and product -6. Thus, there are exactly three such possibilities.

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