For the integer $n$, if $n^3$ is odd, then which of the following statements are true? I. $n$ is odd. II. $n^2$ is odd. III. $n^2$ is even.

### Concept & Logic The parity of an integer is preserved when raised to a positive integer power. This means if a number is odd, any positive power of that number will be odd, and conversely, if a positive power of an integer is odd, the base integer itself must be odd. ### Step-by-Step Solution **Given:** * $n$ is an integer. * $n^3$ is an odd number. **Calculation / Deduction:** First, determine the parity of $n$ based on $n^3$. The cube of an even number is even (e.g., $2^3 = 8$). The cube of an odd number is odd (e.g., $3^3 = 27$). Since $n^3$ is odd, the base $n$ MUST be an odd integer. * This makes **Statement I ($n$ is odd)** True. Next, evaluate $n^2$. Since we established that $n$ is odd, $n^2$ is the product of two odd numbers (Odd $\times$ Odd). The product of two odd numbers is always odd (e.g., $3^2 = 9$). * This makes **Statement II ($n^2$ is odd)** True. * Consequently, **Statement III ($n^2$ is even)** must be False. Statements I and II are the only true statements. ### Exam Strategy & Shortcut Pick a small odd perfect cube, like $27$. Then $n = 3$. Check the statements rapidly: $n = 3$ (odd - Statement I is true), $n^2 = 9$ (odd - Statement II is true, Statement III is false). Both I and II are verified in under 5 seconds. ### Common Pitfall A common mistake is confusing the rules of parity for addition with those for multiplication, leading students to mistakenly believe that squaring an odd number makes it even. Always fall back to testing with the number 3 if unsure. ### Final Answer Therefore, the correct answer is **I and II only**.