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But if x is 1 won't give that answer as 0?

houston1980
Option D is 3 consecutive integers. Since x is a positive integer, the minimum value of x is 1.
Therefore option D will always be divisible by 3.
Hence D = (x−1)(x)(x+1) is the answer.
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RahulPaul
But if x is 1 won't give that answer as 0?



Yes, when x = 1, option D, (x - 1)(x)(x + 1), equals 0. However, recall that 0 is divisible by every integer except 0 itself.


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 (\(\frac{0}{x} = 0\), 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. \(0! = 1! = 1\).


Hope it helps.
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