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(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. _________________

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

Show Tags

09 Apr 2011, 20:05

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
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

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

Show Tags

07 Sep 2011, 02:59

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
_________________

"The best day of your life is the one on which you decide your life is your own. No apologies or excuses. No one to lean on, rely on, or blame. The gift is yours - it is an amazing journey - and you alone are responsible for the quality of it. This is the day your life really begins." - Bob Moawab

gmatclubot

Re: If x is positive is x divisible by 2
[#permalink]
07 Sep 2011, 02:59

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