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Re: If n is an integer, is (n - 1)(n + 1) a multiple of 24? [#permalink]
Bunuel wrote:
VibhuAnurag wrote:
The answer should be (E).

(1) n is odd:
Let n = 1 --> (n - 1)(n + 1) = 0*2 --> No
Let n = 23 --> (n - 1)(n + 1) = 22*24 --> Yes
Conflicting answers, hence, insufficient.

(2) n is not divisible by 3 (we can take the same numbers again for the sake of simplicity):
Let n = 1 --> (n - 1)(n + 1) = 0*2 --> No
Let n = 23 --> (n - 1)(n + 1) = 22*24 --> Yes
Conflicting answers, hence, insufficient.

Combined:
Since no new information is available, still insufficient.


0 is a multiple of every integer.

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


Noted and edited the solution. Thanks for the correction.
GMAT Club Bot
Re: If n is an integer, is (n - 1)(n + 1) a multiple of 24? [#permalink]
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