If x is positive is x divisible by 2 ? (1) x^3+x is : GMAT Data Sufficiency (DS)
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# If x is positive is x divisible by 2 ? (1) x^3+x is

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If x is positive is x divisible by 2 ? (1) x^3+x is [#permalink]

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22 Mar 2010, 10:56
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If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
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Re: If x is positive is x divisible by 2 [#permalink]

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22 Mar 2010, 12:01
IMO B

A) x could be 2 or 3

B) x is 4, 10, 16 so it jumps every 6 numbers and even plus even equals even so x will always be even therefore divisible by 2. What's the OA?
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Re: If x is positive is x divisible by 2 [#permalink]

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22 Mar 2010, 13:55
It can't be B. 5x+4=6,12,18,24, 30 ... solving those for x give some values of x that are divisible by 2, and some that are not.
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Re: If x is positive is x divisible by 2 [#permalink]

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22 Mar 2010, 17:39
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anaik100 wrote:
If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6

I don't think that this is a GMAT question. In it's current form it's way beyond the GMAT scope.

First of all note that we are not told that $$x$$ is an integer.

Given: $$x>0$$. Question: is $$x$$ even (integer)?

(1) $$x^3+x=x(x^2+1)=4n$$

If x is an integer then either:
A. $$x$$ is a multiple of 4 and $$x^2+1$$ is odd;
OR
B. $$x$$ is odd and $$x^2+1$$ is a multiple of 4. But this scenario is not possible. Why? Because in case $$x=2k+1$$ ($$x$$ is odd), $$x^2+1=4k^2+4k+2$$, which is not a multiple of 4, (it's even never multiple of 4.)
So if $$x$$ is an integer it must be multiple of 4.

But $$x$$ also can be a non-integer, for example equation $$x^3+x=4$$ has one real root ($$x\approx{1.38}$$), which is not an integer OR equation $$x^3+x=8$$ also has one real root ($$x\approx{1.83}$$), which is not an integer.

Not sufficient.

(2) $$5x+4=6k$$

If $$x$$ is an integer, then $$x$$ must be even as $$5x+4=even$$ --> $$5x=even-4=even$$ --> $$x=even$$.

But $$x$$ can also be a fraction, for example $$\frac{2}{5}$$, or $$\frac{8}{5}$$ - basically $$x$$ can be a fraction of a form $$x=\frac{a}{5}$$, where $$a$$ is an integer.

Not sufficient.

(1)+(2) Now if $$x$$ is an integer then it's even as we concluded.

But can x be reduced fraction of a form $$x=\frac{a}{5}$$? We'll have $$\frac{a^3}{125}+\frac{a}{5}=4n$$ --> $$a^3+25a=a(a^2+25)=500n$$ --> as $$a$$ and $$n$$ are integers, $$a$$ must be multiple of 5 (at least) --> hence $$x=\frac{a}{5}=integer$$, so $$x$$ can not be reduced fraction --> and if $$x$$ is an integer, then we know it must be even.

Answer: C.

If we were told that $$x$$ is an integer, then the answer would be D.
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is x divisible by 2 ? [#permalink]

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23 May 2010, 20:34
If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6
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Re: is x divisible by 2 ? [#permalink]

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23 May 2010, 22:15
Ans B.

A - x(x^2 +1)/4. whether x or x^2+1 is divisible by 4. cant say which.
B- (5x + 4)/6. x= 4, 16, ....divisible by 2.
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Re: If x is positive is x divisible by 2 [#permalink]

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09 Apr 2011, 17:57
Answer C
Statement (2) alone is sufficient to answer.

Explaination -
For (1) X = Sqrt(3) or X=4

For (2) 5X+4 to be divisible by 6, X has to be an even integer >0 as 5X+4 should be even. So X will always be
divisible by 2

Oh Yes. 2 is not sufficient. Missed that X can be a fraction with a denominator of 5.

Answer c.
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Re: If x is positive is x divisible by 2 [#permalink]

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09 Apr 2011, 19:05
If we can assume that x is an integer :

Is x even ?

From (1)

x^3 + x = 4k

x(x^2+1) = 4k

if x = Odd, then X^2 + 1 = even and is divisible by 2, but x is not divisible by 4, and X^2 + 1 needs one more 2 in its factor, which is not possible it it's odd. So x is even.

(Sufficient)

From (2)

5x + 4 = 6k

=> 5x + 4 has 2 and 3 as factors

So, 5x + 4 is even

so 5x + even = even

=> 5x = even

=> x = even

(sufficient)

Answer - D
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Re: If x is positive is x divisible by 2 [#permalink]

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10 Apr 2011, 00:31
fluke,
I don't think above posters are correct. sqrt(3) is not a valid data for S1.
1.732*(4) is not divisible by 4.

1) x = 4. The answer is YES
x = 12 . The answer is YES
Sufficient.

2) Consider 5x = 14. 5x+4 = 18. The answer is NO
Consider 5x = 20. 5x+4 = 24. The answer is YES
Insufficient.

Hence A
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Re: If x is positive is x divisible by 2 [#permalink]

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07 Sep 2011, 01:59
Quote:
If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4
(2) 5x+4 is divisible by 6

x>0. is x a multiple of 2 (i.e. is x an even number)?

I got B for answer.
I am working based on the assumption that x is an integer, so that the question will be within the scope of GMAT..

Rules applied:
even + even = even
odd + even = odd
odd + odd = even
even * even = even
even * odd = even

Statement 1

x^3+x=4k ---> k being an integer
x(x^2+1)=4k

if x is even, (x^2+1) will be odd, multiplying both results in 4k (even).
if x is odd, (x^2+1) will be even, which also results in 4k (even)
--> insufficient

Statement 2
5x+4=6w ---> w being an integer
if x is even --> 5x will be even --> even + even = even --> results in even outcome i.e. 6w (true)
if x is odd --> 5x will be odd --> odd + even = odd --> results in odd outcome i.e. cannot be 6w
So x must be even.
--> sufficient

Answer: B
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Re: If x is positive is x divisible by 2   [#permalink] 07 Sep 2011, 01:59
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