Let $X_k = (p_1 p_2 \dots p_k) + 1$, where $p_1, p_2, \dots, p_k$ are the first $k$ primes.\n\nConsider the following:\n\n1. $X_k$ is a prime number.\n\n2. $X_k$ is a composite number.\n\n3. $X_k + 1$ is always an even number.\n\nWhich of the above is/are correct?

### Concept & Logic This problem relies on Euclid's proof of the infinitude of primes and basic even/odd number properties. The product of the first $k$ primes, plus one, generates a number that is not divisible by any of those initial primes. ### Step-by-Step Solution * **Analyze Statements 1 & 2:** The formula is $X_k = (p_1 \times p_2 \times \dots \times p_k) + 1$. While $X_k$ is prime for the first few values of $k$ (e.g., $X_1 = 2+1=3$, $X_2 = 2 \times 3 + 1 = 7$), it is not *always* prime. For example, when $k=6$: $$X_6 = (2 \times 3 \times 5 \times 7 \times 11 \times 13) + 1 = 30031$$ $30031$ is composite because $30031 = 59 \times 509$. Because $X_k$ can be prime or composite depending on $k$, both sweeping statements "1. $X_k$ is a prime number" and "2. $X_k$ is a composite number" are false. * **Analyze Statement 3:** The expression is $X_k + 1$. Substitute $X_k$: $$X_k + 1 = (p_1 \times p_2 \times \dots \times p_k) + 1 + 1$$ $$X_k + 1 = (p_1 \times p_2 \times \dots \times p_k) + 2$$ Since the first prime $p_1$ is always 2, the product $(p_1 \dots p_k)$ is always an even number. $$\text{Even Number} + 2 = \text{Even Number}$$ Therefore, $X_k + 1$ is unconditionally an even number. Statement 3 is correct. ### Exam Strategy & Shortcut You don't need to memorize the $X_6$ exception to solve this. Because options 1 and 2 make absolute claims that contradict each other, suspect they are both conditionally true and therefore false as general rules. Focus on Statement 3. You know $p_1 = 2$, so the product block is even. $X_k$ is (Even + 1) = Odd. Thus, $X_k + 1$ is (Odd + 1) = Even. This guarantees Statement 3 is correct. ### Common Pitfall Confusing the equation in Euclid's proof with a "prime generating formula". Euclid's proof merely shows $X_k$ has prime factors *larger* than $p_k$, it does not guarantee $X_k$ itself is prime. Assuming Statement 1 is true is the most common error. ### Final Answer **Therefore, the correct answer is 3 only.**