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Comment on Is x Even
For statement 2, I plugged in
Can you show me your
Can you show me your calculations. I ask, because I think your calculations might be off.
Statement 2: 6x - 3y is odd
- case a: if x is even and y is odd, then 6x - 3y is ODD. In this case, x IS even
- case b: if x is odd and y is odd, then 6x - 3y is ODD. In this case, x is NOT even
Both cases satisfy statement 2, but each case yields a different answer to the target question. This means statement 2 is not sufficient.
Hi Brent, in case b: if x is
But we are getting -12 when it should be odd? That's why I got St 2 insufficient. What's missing here? Thanks Brent
Sorry, I made a small mistake
Sorry, I made a small mistake in my earlier post (which I have now corrected).
case b should read: "if x is odd and y is ODD, then 6x - 3y is ODD.
Thanks Brent.
can statement 2 be solved
Be careful. Statement 2 does
Be careful. Statement 2 does talk about the difference between x and y; it talks about the difference between 6x and 3y. Since 6x must be EVEN, we can conclude that y is odd.
If the target question had asked "Is y odd?", your approach would have found statement 2 to be insufficient, when it would have been sufficient.
For statement 1, if I look at
Your first statement is true.
Your first statement is true.
However, your second statement is not true: Alternatively, if y=even, then x will have to be odd since first term has to be odd.
If y is even, then the first term (xy) will be even, regardless of the value of x.
This tells us that, if xy+y is odd, then y must be odd.
Cheers,
Brent
for s2 you can just factor
we happen to know 2x is always odd so we don't know if x is odd or not so not sufficient
Be careful, 2x is always EVEN
Be careful, 2x is always EVEN (not odd) for all integer values of x.
You're correct to say that we can factor statement 2 to get: 3(2x - y) is ODD.
Since (ODD)(ODD) = ODD, we know that (2x - y) is ODD.
case i: It could be the case that x = 1 and y = 1, in which case x is ODD.
case ii: It could also be the case that x = 2 and y = 1, in which case x is EVEN.
So, statement 2 is not sufficient.