if | r | != 1 is integer r even 1 r is not positive 2 2r

First of all factorial of a number equals to 1 in two cases: number equals to 0 --> 0!=1 or number equals to 1 --> 1!=1.

\(|r|!=1\) --> \(|r|=0\) or \(|r|=1\), so \(r\) can take 3 values: 0, 1, and -1, out of which only 0 is even. So basicalyy questions asks: is \(r=0\)?

(1) \(r\) is not positive --> \(r\) can be 0 or -1. Not sufficient.

(2) \(2r>-5\) --> \(r>-\frac{5}{2}=-2.5\) --> \(r\) can be 0, 1 or -1. Not sufficient.

(1)+(2) still two values of \(r\) are possible 0 and -1. Not sufficient.

Answer: E.

If statement (2) were \(5r>-2\), then answer would be C: \(r>-\frac{2}{5}=-0.4\) --> \(r\) can be 0 or 1. Not sufficient.

(1)+(2) only one value of \(r\) is possible: 0, hence \(r\) is even. Sufficient.
_________________

Bunuel u mistook the statement

what it means is | r | != 1 i meant | r | not equal to 1 != ( not equal) in programming world i am sorry for this confusion

This means that I wrote new question:

If r is an integer and \(|r|!=1\) is \(r\) even?

(1) \(r\) is not positive.

(2) \(5r>-2\).

Answer: C.

As for original question:

If \(|r|\neq{1}\) is \(r=even\)?

\(|r|\neq{1}\) --> \(r\neq{1}\) and \(r\neq{-1}\).

(1) \(r\) is not positive --> Clearly insufficient, \(r\) can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) \(2r>-5\) --> \(r>-\frac{5}{2}=-2.5\) --> again \(r\) can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.

(1)+(2) \(r\) is not positive and \(r>-2.5\) --> \(r\) can be -2, -1, or 0. But as given that \(r\neq{-1}\) then only valid solutions for \(r\) are -2 and 0, both are even. Sufficient.

Answer: C.
_________________

Bunuel,
If condition (2) above is correct = 5r>-2, it would mean r =0,1,2,.. I think you have the condition flipped?
and if r =0,1,2, and from (1) we have r <> 1, <> -1, I believe that cannot let us confirm that r is even. So E

I just saw, (2) was a typo. Very rare you make a mistake!

In the solution above, is 0 considered positive. If not, then we are left with -2 and that is even, so the answer still works.

true, the condition for statement 2 has been flipped.

No mistake there. Nothing was flipped. Look carefully, I consider in my previous post two questions:

MINE:
If r is an integer and \(|r|!=1\) is \(r\) even?

(1) \(r\) is not positive.

(2) \(5r>-2\).

Answer: C.

AND ORIGINAL:

If \(|r|\neq{1}\) is \(r=even\)?

(1) \(r\) is not positive.

(2) \(2r>-5\).

Answer: C.

Totally in my previous posts we have 3 questions. First 2 I "invented" because rxs0005 meant by | r | != 1 \(|r|\neq{1}\) and I understood it as it was written \(|r|!={1}\). So two questions I consider in my first post are those I "invented".

Question 1 (mine):

If r is an integer and \(|r|!=1\) is \(r\) even?

First of all factorial of a number equals to 1 in two cases: number equals to 0 --> 0!=1 or number equals to 1 --> 1!=1.

\(|r|!=1\) --> \(|r|=0\) or \(|r|=1\), so \(r\) can take 3 values: 0, 1, and -1, out of which only 0 is even. So basically questions asks: is \(r=0\)?

(1) \(r\) is not positive --> \(r\) can be 0 or -1. Not sufficient.

(2) \(2r>-5\) --> \(r>-\frac{5}{2}=-2.5\) --> \(r\) can be 0, 1 or -1. Not sufficient.

(1)+(2) still two values of \(r\) are possible 0 and -1. Not sufficient.

Answer: E.

Question #2 (mine):

If r is an integer and \(|r|!=1\) is \(r\) even?

(1) \(r\) is not positive --> \(r\) can be 0 or -1. Not sufficient.

(2) \(5r>-2\) --> \(r>-\frac{2}{5}=-0.4\) --> \(r\) can be 0 or 1. Not sufficient.

(1)+(2) only one value of \(r\) is possible: 0, hence \(r\) is even. Sufficient.

Answer: C.

Question #3 (original):

If \(|r|\neq{1}\) is \(r=even\)?

\(|r|\neq{1}\) --> \(r\neq{1}\) and \(r\neq{-1}\).

(1) \(r\) is not positive --> Clearly insufficient, \(r\) can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) \(2r>-5\) --> \(r>-\frac{5}{2}=-2.5\) --> again \(r\) can be even or odd (except -1 and 1): -2, 0, 2, 3, 4, 5, ... Not sufficient.

(1)+(2) \(r\) is not positive and \(r>-2.5\) --> \(r\) can be -2, -1, or 0. But as given that \(r\neq{-1}\) then only valid solutions for \(r\) are -2 and 0, both are even. Sufficient.

Answer: C.

BUT: if the second statement were as you wrote answer still would be C.

Question #4 (your's):

If \(|r|\neq{1}\) is \(r=even\)?

\(|r|\neq{1}\) --> \(r\neq{1}\) and \(r\neq{-1}\).

(1) \(r\) is not positive --> Clearly insufficient, \(r\) can be any non-positive number (except -1) even or odd (0, -2, -3, -4, ...).

(2) \(5r>-2\) --> \(r>-\frac{2}{5}=-0.4\) --> \(r\) can be 0 or 2, 3, 4, ... Not sufficient.

(1)+(2) Only one solution is possible for \(r\) is 0, which is even. Sufficient.

Answer: C.

Hope it's clear.
_________________

Bunuel - I see that there was an original question and you had modified it. I got it. Thanks for the clarification. As I said, very rare you make a mistake

OA on original question should be E.
Please correct

Thanks

Cheers!
J

Question formatting done.
OA is correct
_________________

Hi jlgdr,
The answer for the original question is indeed E , if "!" meant as factorial. However if the same is mentioned as "not equal to", then the answer should be C. I believe the question is originally posted for "!" as nt equal to. In that case "C" is the correct answer as mentioned in the OA.
Hope that helps.

Thanks.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________