If $n = 1 + x$, where $x$ is the product of four consecutive positive integers, then which of the following is/are true? I. $n$ is odd. II. $n$ is prime. III. $n$ is a perfect square.

### Concept & Logic This is an advanced number property theorem. The product of any four consecutive positive integers plus one is ALWAYS a perfect square. Additionally, understanding the parity (odd/even nature) of consecutive integers solves the first statement. ### Step-by-Step Solution * **Given:** $x = k(k+1)(k+2)(k+3)$, where $k$ is a positive integer. $n = x + 1$. * Let's evaluate the statements: * **Statement I ($n$ is odd):** In any sequence of four consecutive integers, two must be even. The product of integers where at least one is even results in an even product. Thus, $x$ is always even. Since $n = x + 1$, adding $1$ to an even number makes $n$ definitively **odd**. Statement I is true. * **Statement II ($n$ is prime):** Let's test with the smallest positive integers. If $k=1$, $x = 1 \times 2 \times 3 \times 4 = 24$. Therefore, $n = 24 + 1 = 25$. $25$ is divisible by $5$, so it is not prime. Statement II is false. * **Statement III ($n$ is a perfect square):** Let's prove the theorem algebraically. $$x = k(k+1)(k+2)(k+3)$$ Group the outer and inner terms: $$x = [k(k+3)] \times [(k+1)(k+2)]$$ $$x = (k^2 + 3k)(k^2 + 3k + 2)$$ Let $u = k^2 + 3k$. The expression becomes: $$x = u(u + 2) = u^2 + 2u$$ Now, calculate $n$: $$n = x + 1 = (u^2 + 2u) + 1$$ Notice this is a perfect trinomial square: $$n = (u + 1)^2$$ Since $u$ is an integer, $n$ is inherently a perfect square. Statement III is true. ### Exam Strategy & Shortcut **Small Integer Testing:** You do not need to derive the algebraic proof during an exam. Test the smallest possible case: $1 \times 2 \times 3 \times 4 = 24$. Then $n = 24 + 1 = 25$. Is $25$ odd? Yes. Is $25$ prime? No. Is $25$ a perfect square? Yes ($5^2$). The only option containing just I and III is the correct answer. ### Common Pitfall Assuming that a complex algebraic definition for a sequence will inherently generate primes. Students often guess that such unique mathematical structures yield prime numbers without testing the base case. ### Final Answer Therefore, the correct answer is **I and III only**.