For integers x and y , x=y/6 +1 . Is xy even?

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by neelgandham » Thu Oct 27, 2011 12:43 pm
For integers x and y , x=y/6 +1 => y =6x-1 (if x=even, y=odd and x=odd, y =odd). Is xy even?

(1) y+1/5 =x

=> y=5x-1, if x=even, y=odd and x=odd, y =even. from the condition y =6x-1 we already have x=even, y=odd, so we know that x is even and y is odd !

(2) y/x is not odd => y/x = even => y =even*x => y =even => xy=even as y is even

Note: 0 is an even number

Answer:D.
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by HSPA » Fri Nov 11, 2011 8:06 pm
Is your question correct...it has to be B only.. because for integer X and Y the two equations wont satisfy

x= y/6+1 and x = y+1/5

Your question stands for assumption: x = (y+1)/5 and fails for x = y+ 1/5
lenagmat wrote:For integers x and y , x=y/6 +1 . Is xy even?

(1) y+1/5 =x
(2) y/x is not odd

D.
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by user123321 » Fri Nov 11, 2011 9:19 pm
I am not so sure about choice D. IMO A

if y and x are integers, if they say y/x is not odd, it doesn't mean that y/x is even, it can be decimal as well(say x = 2, y =1 => y/x = 0.5 not an odd number).

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by HSPA » Fri Nov 11, 2011 10:26 pm
Awesome.. I agree for A ( my assumption included) But my math knowledge says odd and even are only for integers, user123321 have found one more assumption... if not odd then even ( only binary)..y/x shall not have a fraction is implicit assumption.
user123321 wrote:I am not so sure about choice D. IMO A

if y and x are integers, if they say y/x is not odd, it doesn't mean that y/x is even, it can be decimal as well(say x = 2, y =1 => y/x = 0.5 not an odd number).

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