Ok, another question, hopefully I won't correct myself again..

Stuart Kovinsky wrote:

naveedakhan wrote:2x can either be even or odd, doesn't matter, according the statement 1, we only take the values of x for which the compete statement (2x-1) is odd.
The only reason of ruling out A would be if x=0 because 2(0)-1=-1 which is odd but that x=0 is even, otherwise for any other values, either fractions or negatives, as long as 2x-1 churns out an odd number, the value for x can be taken as such.

Stuart...please comment

2x MUST be even according to statement (1).

If 2x - 1 = odd, then we can say that:

2x = odd + 1

and any odd number + 1 will give you an even number.

However, just knowing that 2x is even doesn't help us decide if x is odd or even, it only tells us that x is an integer.

2x = even
x = even/2

and every even number is divisible by 2, so we know that x must be an integer.

Does one not have to consider the possibility that x=0 because 2(0)-1=-1? I understand the thinking behind 2x = odd + 1 = even (and I would love to just leave it at that), but as you said yourself: never make assumptions. So why assume that x is not 0?

Thanks in advance,
Simon

sdilmanian wrote:Ok, another question, hopefully I won't correct myself again..

Stuart Kovinsky wrote:

naveedakhan wrote:2x can either be even or odd, doesn't matter, according the statement 1, we only take the values of x for which the compete statement (2x-1) is odd.
The only reason of ruling out A would be if x=0 because 2(0)-1=-1 which is odd but that x=0 is even, otherwise for any other values, either fractions or negatives, as long as 2x-1 churns out an odd number, the value for x can be taken as such.

Stuart...please comment

2x MUST be even according to statement (1).

If 2x - 1 = odd, then we can say that:

2x = odd + 1

and any odd number + 1 will give you an even number.

However, just knowing that 2x is even doesn't help us decide if x is odd or even, it only tells us that x is an integer.

2x = even
x = even/2

and every even number is divisible by 2, so we know that x must be an integer.

Does one not have to consider the possibility that x=0 because 2(0)-1=-1? I understand the thinking behind 2x = odd + 1 = even (and I would love to just leave it at that), but as you said yourself: never make assumptions. So why assume that x is not 0?

Thanks in advance,
Simon

Hi,

there's absolutely no reason why 2x couldn't be 0; after all, 0 is even.

However, considering x=0 for statement (1) doesn't change the answer to the question.

Remember, we need to get a "definite yes" or "definite no" for a statement to be sufficient. Let's look at (1) again:

(1) 2x - 1 is odd

Writing that as an equation:

2x - 1 = odd
2x = odd + 1
2x = even
x = even/2

Could x = 0? Sure, since 0 = 0/2 (and since 0 is an even integer).
If x=0, is x odd? NO.

However, x could also equal 1, since 1 = 2/2 (and since 2 is an even integer).
If x=1, is x odd? YES.

(In fact, if we did the math, we'd see that if 2x is a multiple of 4, e.g. {... -4, 0, 4, 8, 12, ...} x is always even; if 2x is not a multiple of 4, e.g. {-6, -2, 2, 6, 10, ...} x is always odd.)

Since we can get both a yes and a no answer, (1) is insufficient alone.

Is x odd?

1) 2x-1 is odd
2) x^3 is odd

a) 2x-1 is odd or 2x is even. x=2 an x=1 both satisfy this. So insufficient
b) x^3 is odd. but we don't know if x is integer. Insufficient
a&b) from a, x is integer. also x^3 is odd. thus x is odd
IMO C

IMO C

Thanks Stuart. The fact that 0 is officially considered an even integer wasn't clear to me, hence my confusion. But this definitely cleared it up.
Thanks!