UPSC CDS II 2025 Mathematics Paper

Concept:For an even $n$, the product $n(n+1)(n+2)$ always contains a factor of $8$ and a factor of $3$, so it is always divisible by $24$. The square then inherits a larger divisor.Explanation:Let $n = 2k$, where $k$ is a natural number.Then $x = 2k(2k+1)(2k+2) = 2k(2k+1)\cdot2(k+1) = 4k(k+1)(2k+1)$.Among $k$ and $k+1$, one is even, so their product is divisible by $2$. Hence $x$ has factor $4 \times 2 = 8$.Among any three consecutive integers $k$, $k+1$, $2k+1$, one is a multiple of $3$. Therefore $x$ is always divisible by $8 \times 3 = 24$.Check Statement I: $48 = 16 \times 3$. But $x$ only guarantees $8 \times 3$. For example, $n=2$ gives $x=2\times3\times4=24$, which is not divisible by $48$. So Statement I is false.Check Statement II: Since $x$ is always divisible by $24$, $x^2$ is divisible by $24^2 = 576$. Now $144 = 16 \times 9$ divides $576$, so $x^2$ is always divisible by $144$. Thus Statement II is true.Answer:B. II only