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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]

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

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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
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