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20 May 2024, 01:30
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If \(x\) is a positive integer and the difference between any two distinct positive factors of \(x\) is even, which of the following must be true?
I. \(x^2 - 1\) is even.
II. \(\frac{x+1}{2}\) is even.
III. \(x\) is not a prime number.
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
I. \(x^2 - 1\) is even.
II. \(\frac{x+1}{2}\) is even.
III. \(x\) is not a prime number.
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
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02 Jun 2024, 15:15
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Official Solution:
If \(x\) is a positive integer and the difference between any two distinct positive factors of \(x\) is even, which of the following must be true?
I. \(x^2 - 1\) is even.
II. \(\frac{x+1}{2}\) is even.
III. \(x\) is not a prime number.
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
Given that the difference of any two distinct positive factors of \(x\) is even, \(x\) cannot be even because if it is, then it has 1 and 2 as factors and \(1 - 2 = 1 = odd\). Thus, \(x\) must be odd. In this case, all its factors will be odd and the difference of any two distinct positive factors of \(x\) will be even.
If \(x\) is odd, then \(x^2 - 1 = \text{odd}^2 - 1 = \text{odd} - 1 = \text{even}\). Thus, I must be true. To discard the other two options, consider \(x = 5\).
Answer: A
If \(x\) is a positive integer and the difference between any two distinct positive factors of \(x\) is even, which of the following must be true?
I. \(x^2 - 1\) is even.
II. \(\frac{x+1}{2}\) is even.
III. \(x\) is not a prime number.
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
Given that the difference of any two distinct positive factors of \(x\) is even, \(x\) cannot be even because if it is, then it has 1 and 2 as factors and \(1 - 2 = 1 = odd\). Thus, \(x\) must be odd. In this case, all its factors will be odd and the difference of any two distinct positive factors of \(x\) will be even.
If \(x\) is odd, then \(x^2 - 1 = \text{odd}^2 - 1 = \text{odd} - 1 = \text{even}\). Thus, I must be true. To discard the other two options, consider \(x = 5\).
Answer: A
General Discussion
20 May 2024, 01:38
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Bunuel
Quote:
Inference: x is an odd number, as all the factors of x are odd.
Answer choice elimination
I. \(x^2 - 1\) is even.
We know that x is odd. Hence \(x^2\) is also odd.
\(x^2 - 1\) = even
This must be true.
II. \(\frac{x+1}{2}\) is even.
We know that x is odd. Hence, x + 1 = even.
However, if \(x+1\) is even, \(\frac{x + 1}{2}\) may or may not be even.
Ex. \(x = 5\)
\(x + 1 = 6\)
\(\frac{x + 1}{2 } = 3 \) = odd
III. \(x\) is not a prime number.
x can be a prime number. For example, x can be equal to 5.
Option A
