Consider the following statements: 1. If $x$ and $y$ are composite numbers, then $x + y$ is always composite. 2. There does not exist a natural number which is neither prime nor composite. Which of the above statements is/are correct?

### Concept & Logic This problem tests your theoretical understanding of number classifications. To disprove an "always" statement in mathematics, you only need to find a single counter-example. To evaluate existence statements, you must recall boundary cases like the number $1$. ### Step-by-Step Solution Let's evaluate each statement individually. **Statement 1 Evaluation:** * Claim: If $x$ and $y$ are composite, $x + y$ is always composite. * Let's find a counter-example. * Let $x = 4$ (a composite number). * Let $y = 9$ (a composite number). * Calculate the sum: $4 + 9 = 13$. * The sum $13$ is a prime number, not a composite number. * Because we found a case where the sum is prime, Statement 1 is false. **Statement 2 Evaluation:** * Claim: There does not exist a natural number which is neither prime nor composite. * The natural numbers begin at $1, 2, 3, 4, ...$ * By definition, prime numbers have exactly two divisors. Composite numbers have more than two divisors. * The number $1$ has exactly one divisor. Therefore, $1$ is classified as neither prime nor composite. * Because $1$ exists, Statement 2 is false. ### Exam Strategy & Shortcut For Statement 1, quickly mentally add small composite numbers ($4, 6, 8, 9, 10$) together until you hit a prime. $4 + 9 = 13$ is the fastest trigger. For Statement 2, always remember the unique status of the number $1$ when dealing with primes. ### Common Pitfall Students often evaluate Statement 1 by only adding even composite numbers (e.g., $4 + 6 = 10$, $8 + 4 = 12$). Since the sum of two even numbers is always even (and therefore composite if $>2$), they incorrectly assume the statement is true. Always test a mix of even and odd composites. ### Final Answer Therefore, the correct answer is Neither 1 nor 2.