UPSC CDS 1 2023 Math Paper

Concept:We use factorisation and properties of consecutive integers to check divisibility.Explanation:Statement 1: $n^{3} - n = n(n-1)(n+1)$ which is the product of three consecutive integers.It is always divisible by 2 and 3, hence by 6.Statement 2: $n^{5} - n = n(n^{4} - 1) = n(n^{2} - 1)(n^{2} + 1)$.Rewrite as $n(n-1)(n+1)\big[(n^{2} - 4) + 5\big]$.This becomes $(n-2)(n-1)n(n+1)(n+2) + 5n(n-1)(n+1)$.The first term is the product of five consecutive integers, always divisible by 5.The second term is also divisible by 5. Hence $n^{5} - n$ is divisible by 5.Statement 3: $n^{5} - 5n^{3} + 4n = n(n^{4} - 5n^{2} + 4)$.Factor: $n(n^{2} - 4)(n^{2} - 1) = (n-2)(n-1)n(n+1)(n+2)$.This product of five consecutive integers is always divisible by 5! = 120.All three statements are correct.Answer:Option D: 1, 2 and 3 are correct.