If X is a natural number, is X+6 odd?
Statements:
1. X−15 is a whole number.
2. X−6 is an odd number.
The question asks whether $X+6$ is an odd number, given that $X$ is a natural number.
Statement 1 Analysis
Statement 1 says $X-15$ is a whole number.
- Since $X$ is a natural number, $X \geq 1$.
- Whole numbers are $0, 1, 2, ...$. Thus, $X-15 \geq 0$.
- This implies $X \geq 15$.
- If $X=15$ (a natural number $\geq 15$), then $X+6 = 15+6 = 21$, which is odd.
- If $X=16$ (a natural number $\geq 15$), then $X+6 = 16+6 = 22$, which is even.
- Since $X+6$ can be either odd or even depending on the value of $X$, Statement 1 alone is not sufficient.
Statement 2 Analysis
Statement 2 says $X-6$ is an odd number.
- Let $X-6 = k$, where $k$ is an odd number.
- This means $X = k+6$.
- We need to determine the parity of $X+6$.
- Substituting $X$: $X+6 = (k+6) + 6 = k + 12$.
- Since $k$ is odd and $12$ is even, their sum ($k+12$) must be odd.
- Therefore, $X+6$ is always odd.
- Statement 2 alone is sufficient.
Conclusion
Statement 1 alone is not sufficient, but Statement 2 alone is sufficient to answer the question.
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