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M08-24

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:37
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45% (medium)

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58% (00:40) correct 42% (00:48) wrong based on 124 sessions

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If $$x$$ is an odd integer, which of the following numbers must be even?

A. $$x^{\frac{3}{3}}$$
B. $$\frac{x^2 - 1}{2}$$
C. $$xp + 1$$ where $$p$$ is a prime number
D. $$\frac{(x + 7)(x - 2)}{2}$$
E. $$x^{x - 1}$$

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16 Sep 2014, 00:37
2
Official Solution:

If $$x$$ is an odd integer, which of the following numbers must be even?

A. $$x^{\frac{3}{3}}$$
B. $$\frac{x^2 - 1}{2}$$
C. $$xp + 1$$ where $$p$$ is a prime number
D. $$\frac{(x + 7)(x - 2)}{2}$$
E. $$x^{x - 1}$$

$$\frac{(x^2 - 1)}{2}$$ can be written as $$(x - 1)*\frac{(x + 1)}{2}$$ = even*even/2 = even*integer = even. To see that other choices are not necessarily even consider $$x = 3$$ (in $$C$$ consider $$p = 2$$).

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10 Feb 2016, 04:36
but what if i choose x=1 then answer choice B would be 0, which is not even
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10 Feb 2016, 04:47
1
nausherwan wrote:
but what if i choose x=1 then answer choice B would be 0, which is not even

Hi,
0 is an even number...
what 0 is not is- it is neither positive nor negative..

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25 Jul 2016, 12:43
I understand the solution,, but what is wrong with this understanding

(x+7)(x−2)/2 => x+7 = odd+odd = even
x-2 => odd -even = odd
(x+7)(x−2) => even*odd = even
even/2 (even) => even
what is wrong with above ?
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25 Jul 2016, 12:46
1
mmrigank wrote:
I understand the solution,, but what is wrong with this understanding

(x+7)(x−2)/2 => x+7 = odd+odd = even
x-2 => odd -even = odd
(x+7)(x−2) => even*odd = even
even/2 (even) => even
what is wrong with above ?

even/2 is not always even. For example, 2/2 = 1 = odd.
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27 Aug 2016, 23:54
Bunuel wrote:
mmrigank wrote:
I understand the solution,, but what is wrong with this understanding

(x+7)(x−2)/2 => x+7 = odd+odd = even
x-2 => odd -even = odd
(x+7)(x−2) => even*odd = even
even/2 (even) => even
what is wrong with above ?

even/2 is not always even. For example, 2/2 = 1 = odd.

Bunuel,
Not matter what is the value of highlighted part, since x+7 is even , then solution will always be even.
Also, what odd value of x will give (x-2) as 2 . It is given that x is odd integer.
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28 Aug 2016, 05:37
nipunjain14 wrote:
Bunuel wrote:
mmrigank wrote:
I understand the solution,, but what is wrong with this understanding

(x+7)(x−2)/2 => x+7 = odd+odd = even
x-2 => odd -even = odd
(x+7)(x−2) => even*odd = even
even/2 (even) => even
what is wrong with above ?

even/2 is not always even. For example, 2/2 = 1 = odd.

Bunuel,
Not matter what is the value of highlighted part, since x+7 is even , then solution will always be even.
Also, what odd value of x will give (x-2) as 2 . It is given that x is odd integer.

It's not clear what are you trying to say...

(x+7)(x−2)/2 = even*odd/2 = even/2, which may or may not be even. Consider x = 3 and x = 5.
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02 Dec 2016, 17:52
Hi Bunuel, why would we use different values for x? If we consider that x is an odd integer (either 3,5,1 etc) option three would always be true. I think that's the point mmrigank was trying to make.

If the question only mentions that x is an odd integer (and there're no quadratics involved), shouldn't we assume that x has only one value?
Thanks!
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30 Jul 2017, 07:58
what is wrong with my understanding ??

Since X is odd , X ^ 2 will always be odd.

Now , (Odd - 1 ) = Even ....

Even / 2 = Could be ODD or EVEN ( Eg : 12/2 or 14/2)

So the expression cannot be even always....so must be even fails???
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30 Jul 2017, 11:36
vishalkumar4mba wrote:
what is wrong with my understanding ??

Since X is odd , X ^ 2 will always be odd.

Now , (Odd - 1 ) = Even ....

Even / 2 = Could be ODD or EVEN ( Eg : 12/2 or 14/2)

So the expression cannot be even always....so must be even fails???

odd^2 - 1 is always even, yes but it's not simply even but a multiple of 4.

1^1 - 1 = 0 = a multiple of 4;
3^1 - 1 = 8 = a multiple of 4;
5^1 - 1 = 24 = a multiple of 4;
7^1 - 1 = 48 = a multiple of 4;
...

So, (odd^2 - 1)/2 = (a multiple of 4)/2 = even.

Hope it's clear.
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Re: M08-24   [#permalink] 30 Jul 2017, 11:36
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