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

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if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]  14 Aug 2010, 03:38
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Question Stats:

42% (02:01) correct 58% (00:51) wrong based on 79 sessions
Orginiallly Posted Question

if | r | != 1 is integer r even

1 r is not positive

2 2r > -5

OA is C but i am not convinced

CORRECT QUESTION

If r is an integer and $$|r|!=1$$ is $$r$$ even?

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

(2) $$5r>-2$$.
[Reveal] Spoiler: OA

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Last edited by WoundedTiger on 16 Apr 2014, 00:13, edited 1 time in total.
Question Formatting done
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Re: Number Prop DS [#permalink]  14 Aug 2010, 04:18
Expert's post
rxs0005 wrote:
if | r | != 1 is integer r even

1 r is not positive

2 2r > -5

OA is C but i am not convinced

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.

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.
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Re: Number Prop DS [#permalink]  14 Aug 2010, 06:53
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
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Re: Number Prop DS [#permalink]  14 Aug 2010, 07:15
Expert's post
rxs0005 wrote:
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) $$5r>-2$$.

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.

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Re: Number Prop DS [#permalink]  14 Aug 2010, 07:33
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.
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Re: Number Prop DS [#permalink]  14 Aug 2010, 07:55
true, the condition for statement 2 has been flipped.
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Re: Number Prop DS [#permalink]  14 Aug 2010, 08:07
Expert's post
mainhoon wrote:
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.

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) $$5r>-2$$.

AND ORIGINAL:

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

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

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

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.

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.

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.

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.

Hope it's clear.
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Re: Number Prop DS [#permalink]  14 Aug 2010, 08:22
Bunuel wrote:
rxs0005 wrote:
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) $$5r>-2$$.

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.

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
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]  15 Apr 2014, 13:01
OA on original question should be E.

Thanks

Cheers!
J
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]  16 Apr 2014, 00:14
jlgdr wrote:
OA on original question should be E.

Thanks

Cheers!
J

Question formatting done.
OA is correct
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Re: if | r | != 1 is integer r even 1 r is not positive 2 2r [#permalink]  22 Apr 2014, 05:57
jlgdr wrote:
OA on original question should be E.

Thanks

Cheers!
J

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.
Re: if | r | != 1 is integer r even 1 r is not positive 2 2r   [#permalink] 22 Apr 2014, 05:57
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