divisibility

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divisibility

by mehrasa » Sat Nov 19, 2011 9:09 pm
if x is positive integer, is x divided by 2?

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

to me the answer is B but OA is D.. could someone explain
Source: — Data Sufficiency |

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by shankar.ashwin » Sat Nov 19, 2011 9:39 pm
WE need to find if x is even here.

Statement 1:

x*(x^2+1) = multiple of 4.

Consider 2 cases of 'x' being odd or even.

'x' is odd, then x^2+1 will be even, but it will have only one factor of 2 (or) this case cannot be possible. Try numbers (x=3(10), x=5(26) , x=11(122) ) and so on. We can see none of these numbers will be divisible by 4. SO x cannot be odd. Therefore, without trying the case of 'x' is even, we can tell it will be even. Sufficient.

Statement 2:

5x + 6 divisible by 6 -> 5x is divisible by 6. Therefore 'x' is even again, Sufficient.

D IMO

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by pemdas » Sun Nov 20, 2011 2:37 am
if x>=1 (x is integer), is x/2 an integer?
st(1) x(x^2+1) is divisible by 4 -> Case A)if x is odd, the odd squared is odd and added by the other odd must be even, any even multiplied by an integer (odd or not odd) is even. Since x is odd and (x^2+1) is even, we must be able to divide (x^2+1) by 4. Translating (x^2+1) into (x+1)^2-2x we set a new statement [(x+1)^2 - 2x]/4 OR (x+1)^2 /4 - x/2 which is not quite possible as x is not divisible by 2. Case B)x is even, (x^2+1) must be odd and only x is divisible by 4 suggests we can answer Yes x is divisible by 2 because x is also divisible by 4. Sufficient.

st(2) 5x+6 is divisible by 6 means 5x/6 + 6/6 or we must make sure 5x/6 is an integer which is ONLY possible if x is divisible by 6, as 5 is an odd prime. We answer Yes, x is divisible by 2, because x is also divisible by 6. Sufficient.

d


mehrasa wrote:if x is positive integer, is x divided by 2?

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

to me the answer is B but OA is D.. could someone explain
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