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DS Question -- really confused (Kaplan)

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Source: — Data Sufficiency |
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by clock60 » Sat May 21, 2011 2:15 pm
hi voodoo child
st 1 can be proved with somewhat heavy math, but i am not sure that this way is optimal
(1)x^3+x=4k, where k is +ve integer.one thing to notice that sum of x^3+x results in some even integer that is 4k
we can obtain even integer from: even+even=even. or from odd+odd=even.
if x is even st 1 is suff, let us review what happens if x is odd
if x is odd, it must be perfomed as x=2m+1
here i`ii try to apply concept (a+b)^3=a^3+3a^2b+3ab^2+b^3
(2m+1)^3+(2m+1)=(8m^3+3*4m^2+3*2m+1)+2m+1=8m^3+3*4m^2+8m+2-but this expression is not divisible by 4 as 2 is not divisible by 4without remaider
thus if x is odd, this contradicts st1, and x must be even
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by sourabh33 » Sat May 21, 2011 6:33 pm
IMO D

Given -> X is a +ve Integer

To Find -> Is x divisible by 2 i.e is x even

Evaluating Statement 1

x^3 + x is divisible by 4
x(x^2 + 1) is divisible by 4
Now since x is positive integer, and x^2+1 will be odd (except when x = 1, but then x^3 + x will not be divisible by 4) x has to be a divisible of
And if x is a divisible of 4, x will be a divisible of 2

Therefore Statement 1 is sufficient

we can try by testing numbers (x^3 + x) will be divisible only when x is a multiple of 4
say x = 1 -> (x^3 + x) = 2 therefor not divisible
say x = 3 -> (x^3 + x) = 30 therefor not divisible
say x = 5 -> (x^3 + x) = 130 therefor not divisible
say x = 6 -> (x^3 + x) = 222 therefor not divisible

Evaluating Statement 2

5x+4 is divisible by 6
Now for 5x+4 to be divisible be 5x+4 has to be even as when a number is divisible by another even number it has to be even

Foe 5x+4 to be even 5x has to be even. And for 5x to be even x has to be even

Therefore Statement 2 is also sufficient
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