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Bunuel
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Bunuel
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Dear Bunuel,

The correct answer D is confusing. If possible values of x are (0, 1, 2), then how is number of distinct positive factors of x+2 is a prime number for when x=2?

Many thanks and sorry asking maybe too basic question!
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Dear Bunuel,

The correct answer D is confusing. If possible values of x are (0, 1, 2), then how is number of distinct positive factors of x+2 is a prime number for when x=2?

Many thanks and sorry asking maybe too basic question!

D. Number of distinct positive factors of x + 2 is a prime number.

If x = 2, then x + 2 = 4 and the number of distinct positive factors of 4 is 3 (1, 2, and 4). 3 is a prime number.
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I think this is a high-quality question and I agree with explanation.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Hey

Why is zero not a multiple of itself? I was of the understanding that an integer is always a multiple of itself.

Regards

Gagan

0 IS a multiple of 0. There was a typo in the solution. It should have been: 0 is divisible by every integer except 0 itself, instead it said: "0 is a multiple of every integer instead of 0 itself". Fixed. Thank you!


ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer.

3. Zero is neither positive nor negative (the only one of this kind).

4. Zero is divisible by EVERY integer except 0 itself (\(\frac{x}{0} = 0\), so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer (\(x*0 = 0\), so 0 is a multiple of any number, x).

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 (\(x^0 = 1\))

9. \(0^0\) case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0), \(0^n = 0\).

11. If the exponent n is negative (n < 0), \(0^n\) is undefined, because \(0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}\), which is undefined. You CANNOT take 0 to the negative power.

12. \(0! = 1! = 1\).

So if 0 is neither positive nor negative, how x can be 0 on option C? I don't understand the resolution of this question.
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So if 0 is neither positive nor negative, how x can be 0 on option C? I don't understand the resolution of this question.

And I don't understand your question!

\(|1-x| \lt 2\) implies that m]x[/m] can be 0, 1 or 2.

The question asks, "Which of the following must be true?"

Option C says that x is positive. But if x = 0, this option is not true. Hence, this option is out!
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So if 0 is neither positive nor negative, how x can be 0 on option C? I don't understand the resolution of this question.

And I don't understand your question!

\(|1-x| \lt 2\) implies that m]x[/m] can be 0, 1 or 2.

The question asks, "Which of the following must be true?"

Option C says that x is positive. But if x = 0, this option is not true. Hence, this option is out!

But if 0 is neither positive nor negative, we cannot say that 0 is positive. Therefore, the only options left are 1 and 2, which are positives, therefore option C is true. No?
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So if 0 is neither positive nor negative, how x can be 0 on option C? I don't understand the resolution of this question.

And I don't understand your question!

\(|1-x| \lt 2\) implies that m]x[/m] can be 0, 1 or 2.

The question asks, "Which of the following must be true?"

Option C says that x is positive. But if x = 0, this option is not true. Hence, this option is out!

But if 0 is neither positive nor negative, we cannot say that 0 is positive. Therefore, the only options left are 1 and 2, which are positives, therefore option C is true. No?

No, you're missing the point. From the stem, x can be 0, 1, or 2. The question wants to know which option MUST be true. Option C says x is positive. This isn't always the case because if x = 0, which isn't positive, Option C is NOT true.
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I think this is a high-quality question and I agree with explanation.
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Hi Bunuel, sorry for the basic question but why is D the answer- isn't 0 a factor of all these numbers?
Bunuel
Official Solution:

If \(x\) is an integer and \(|1-x| \lt 2\), then which of the following must be true?

A. \(x\) is not a prime number
B. \(x^2+x\) is not a prime number
C. \(x\) is positive
D. Number of distinct positive factors of \(x+2\) is a prime number
E. \(x\) is not a multiple of an odd prime number


\(|1-x|\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is less than 2:

--(-1)----1----3-- so, \(-1 < x < 3\). Since it is given that \(x\) is an integer, then \(x\) can be 0, 1 or 2.

A. \(x\) is not a prime number. Not true if \(x=2\).

B. \(x^2+x\) is not a prime number. Not true if \(x=1\).

C. \(x\) is positive. Not true if \(x=0\).

D. Number of distinct positive factors of \(x+2\) is a prime number. True for all three values of \(x\).

E. \(x\) is not a multiple of an odd prime number. Not true if \(x=0\), since zero is a multiple of every integer.


Answer: D
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Hi Bunuel, sorry for the basic question but why is D the answer- isn't 0 a factor of all these numbers?
Bunuel
Official Solution:

If \(x\) is an integer and \(|1-x| \lt 2\), then which of the following must be true?

A. \(x\) is not a prime number
B. \(x^2+x\) is not a prime number
C. \(x\) is positive
D. Number of distinct positive factors of \(x+2\) is a prime number
E. \(x\) is not a multiple of an odd prime number


\(|1-x|\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is less than 2:

--(-1)----1----3-- so, \(-1 < x < 3\). Since it is given that \(x\) is an integer, then \(x\) can be 0, 1 or 2.

A. \(x\) is not a prime number. Not true if \(x=2\).

B. \(x^2+x\) is not a prime number. Not true if \(x=1\).

C. \(x\) is positive. Not true if \(x=0\).

D. Number of distinct positive factors of \(x+2\) is a prime number. True for all three values of \(x\).

E. \(x\) is not a multiple of an odd prime number. Not true if \(x=0\), since zero is a multiple of every integer.


Answer: D

0 is not a factor of any integer.
0 is a multiple of every integer.
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I like the solution - it’s helpful.
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