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16 May 2024, 01:07
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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
16 May 2024, 01:07
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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
02 Sep 2024, 23:51
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- **Odd Factors:** The problem states that the difference between any two distinct positive factors of \( x \) is even. If \( x \) were even, it would have 1 and 2 as factors, and their difference (1 - 2 = -1) would be odd. Therefore, \( x \) must be odd. For an odd number, all its factors are odd, and the difference between any two odd numbers is even.
- **Statement I (x^2−1 is even):** Since \( x \) is odd, \( x^2 \) is also odd (because the square of an odd number is odd). Therefore, \( x^2 - 1 \) is even, as subtracting 1 from an odd number results in an even number. Statement I is true.
-
**Statements II and III:** The example \( x = 5 \) (an odd prime) satisfies the condition where the difference between any two distinct positive factors (1 and 5) is even. Thus, Statement II and III do not hold for all cases. Specifically:
- Statement II ((x+1)/2 is even) does not necessarily hold, as shown by \( x = 5 \).
- Statement III (x is not a prime number) is incorrect, as \( x = 5 \) shows that an odd prime can satisfy the conditions.
Thus, the correct answer is I only, which corresponds to option A.
- **Statement I (x^2−1 is even):** Since \( x \) is odd, \( x^2 \) is also odd (because the square of an odd number is odd). Therefore, \( x^2 - 1 \) is even, as subtracting 1 from an odd number results in an even number. Statement I is true.
-
**Statements II and III:** The example \( x = 5 \) (an odd prime) satisfies the condition where the difference between any two distinct positive factors (1 and 5) is even. Thus, Statement II and III do not hold for all cases. Specifically:
- Statement II ((x+1)/2 is even) does not necessarily hold, as shown by \( x = 5 \).
- Statement III (x is not a prime number) is incorrect, as \( x = 5 \) shows that an odd prime can satisfy the conditions.
Thus, the correct answer is I only, which corresponds to option A.
