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Re: M17-27 [#permalink]
I think this is a high-quality question and I agree with explanation. amazing question
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Re: M17-27 [#permalink]
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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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Re M17-27 [#permalink]
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. 0 is an even positive integer ?
Thank for your return
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Yann1402 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. 0 is an even positive integer ?
Thank for your return


Can you please tell me what exactly is unclear in the solution? What is there to elaborate?

ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

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. \(0! = 1! = 1\).
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Re M17-27 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re M17-27 [#permalink]
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