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# Is x odd?

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Updated on: 15 Jul 2012, 04:41
5
00:00

Difficulty:

95% (hard)

Question Stats:

23% (01:19) correct 77% (01:05) wrong based on 116 sessions

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Is x odd?

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

Originally posted by Ellipse on 15 Jul 2012, 04:30.
Last edited by Bunuel on 15 Jul 2012, 04:41, edited 1 time in total.
Edited the question.
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15 Jul 2012, 04:51
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Is x odd?

(1) 2x - 1 is odd --> $$2x-1=odd$$ --> $$2x=even$$ --> $$x=\frac{even}{2}$$ --> $$x$$ is an integer. Now, $$x=\frac{even}{2}$$ could be even (consider $$x=\frac{even}{2}=\frac{4}{2}=2=even$$) as well as odd (consider $$x=\frac{even}{2}=\frac{2}{2}=1=odd$$). Not sufficient.

(2) x^3 is odd. If $$x=integer$$ then in order $$x^3=odd$$ to hold true, it must be odd (answer YES), but $$x$$ could also be some irrational number, for example $$x=\sqrt[3]{5}$$ (answer NO). Not sufficient.

(1)+(2) Since from (1) $$x=integer$$ then from (2) we have that $$x=odd$$. Sufficient.

Hope it's clear.
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16 Jul 2012, 11:18
@Bunuel:

I don't understand your approach for the first argument:

$$2x-1=odd$$

If the result muss be odd, so x must be even. It will be even if x is 2 or greater than 2. If x is odd the result won't be odd. If I say x = 1 so the result will be Zero. But Zero is neither even nor odd. So the satetement is sufficient ?
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16 Jul 2012, 13:06
Alexmsi wrote:
@Bunuel:

I don't understand your approach for the first argument:

$$2x-1=odd$$

If the result muss be odd, so x must be even. It will be even if x is 2 or greater than 2. If x is odd the result won't be odd. If I say x = 1 so the result will be Zero. But Zero is neither even nor odd. So the satetement is sufficient ?

Several things:

1. $$2x-1=odd$$ --> $$2x=odd+1=odd+odd=even$$ --> so $$2x=even$$ --> $$x=\frac{even}{2}=integer$$. Hence $$2x-1=odd$$ just means that $$x$$ is an integer (it can be even as well as odd).

2. If $$x=1$$the result wont be zero, it'l be 1, so odd: $$2*1-1=2-1=1=odd$$.

3. Zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an integer of the form $$n=2k$$, where $$k$$ is an integer. So for $$k=0$$ --> $$n=2*0=0$$.

Hope it's clear.
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04 Feb 2016, 19:09
I understand why A is wrong...for ex. x=0.5...
I understand why B is wrong
but can't see why C is ok...hm..
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04 Feb 2016, 21:31
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mvictor wrote:
I understand why A is wrong...for ex. x=0.5...
I understand why B is wrong
but can't see why C is ok...hm..

Hi,
1) Statement 1 gives you 2x-1 is odd ..
so 2x is even ..
x can be
i) odd,
ii) even or
iii) a fraction with a denominator of 2 in its simplified form..

2) statement 2 gives you x^3 is odd..
this gives you two cases of x..
i) x is odd, or
ii) x is third root of some odd number..

You have correctly realized that A and B are not sufficient alone..

But combined..
for both 1 and 2 to be true, the value of x has to be something that fits in both the cases..
look at the possible values of x, only X IS ODD is common to both..
so x is odd and C is the answer..

Hope it helps you
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11 Oct 2016, 09:01
chetan2u wrote:
mvictor wrote:
I understand why A is wrong...for ex. x=0.5...
I understand why B is wrong
but can't see why C is ok...hm..

Hi,
1) Statement 1 gives you 2x-1 is odd ..
so 2x is even ..
x can be
i) odd,
ii) even or
iii) a fraction with a denominator of 2 in its simplified form..

2) statement 2 gives you x^3 is odd..
this gives you two cases of x..
i) x is odd, or
ii) x is third root of some odd number..

You have correctly realized that A and B are not sufficient alone..

But combined..
for both 1 and 2 to be true, the value of x has to be something that fits in both the cases..
look at the possible values of x, only X IS ODD is common to both..
so x is odd and C is the answer..

Hope it helps you

Sorry, but I think that a fraction with a denominator of 2 in its simplified form is not an option, because then 2x would be odd, and 2x -1= even.
The only thing that statement 1 gives us is that x is an integer.
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21 Nov 2019, 23:22
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Re: Is x odd?   [#permalink] 21 Nov 2019, 23:22
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