Loading...
Is the positive integer \(\mathrm{n}\) the square of an integer?
We need to determine whether the positive integer \(\mathrm{n}\) is a perfect square - that is, can \(\mathrm{n}\) be written as \(\mathrm{k}^2\) for some integer \(\mathrm{k}\)?
For this question to have a sufficient answer, we need to be able to definitively say YES (\(\mathrm{n}\) is always a perfect square) or NO (\(\mathrm{n}\) is never a perfect square) based on the given information.
A number is a perfect square when it equals some integer multiplied by itself (like \(4 = 2 \times 2\), \(9 = 3 \times 3\), \(16 = 4 \times 4\), etc.). We'll use pattern recognition by testing specific examples to discover the underlying rules.
Statement 1: \(4\mathrm{n}\) is the square of an integer
If \(4\mathrm{n}\) is a perfect square, then \(4\mathrm{n} = \mathrm{k}^2\) for some integer \(\mathrm{k}\). Since \(4\) is already a perfect square (\(2^2\)), let's explore what this tells us about \(\mathrm{n}\).
When n IS a perfect square:
When n is NOT a perfect square:
Notice that \(4\mathrm{n}\) is a perfect square only when \(\mathrm{n}\) itself is a perfect square! This makes sense because \(4 = 2^2\), and for the product of two numbers to be a perfect square, both factors need the right "square structure."
Conclusion: Statement 1 guarantees that \(\mathrm{n}\) must be a perfect square.
[STOP - Sufficient!] Statement 1 alone is sufficient - the answer is always YES.
This eliminates choices B, C, and E.
Important: We now forget Statement 1 completely and analyze Statement 2 independently.
Statement 2: \(\mathrm{n}^3\) is the square of an integer
If \(\mathrm{n}^3\) is a perfect square, then \(\mathrm{n}^3 = \mathrm{r}^2\) for some integer \(\mathrm{r}\). Let's explore what happens when we cube different types of numbers.
When n IS a perfect square:
Here's the pattern: when \(\mathrm{n} = \mathrm{m}^2\), then \(\mathrm{n}^3 = (\mathrm{m}^2)^3 = \mathrm{m}^6 = (\mathrm{m}^3)^2\), which is always a perfect square.
When n is NOT a perfect square:
When you cube a number, you're multiplying it by itself three times. For the result to be a perfect square (an even power of all prime factors), the original number must already have the right structure - it must be a perfect square itself.
Conclusion: Statement 2 guarantees that \(\mathrm{n}\) must be a perfect square.
[STOP - Sufficient!] Statement 2 alone is sufficient - the answer is always YES.
Both statements independently guarantee that \(\mathrm{n}\) must be a perfect square. Each statement alone provides sufficient information to answer the question with a definitive YES.
Answer Choice D: Each statement alone is sufficient.