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# If x and y are integers, is y divisible by 2?

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Senior Manager
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If x and y are integers, is y divisible by 2? [#permalink]

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09 Jun 2014, 07:34
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If x and y are integers, is y divisible by 2?

(1) $$x^y=1$$
(2) x is negative.
[Reveal] Spoiler: OA

Kudos [?]: 1791 [1], given: 289

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Re: If x and y are integers, is y divisible by 2? (1) x^y=1 [#permalink]

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09 Jun 2014, 07:51
goodyear2013 wrote:
If x and y are integers, is y divisible by 2?
(1) $$x^y=1$$
(2) x is negative.

X and y are integers,

St.1 says $$x^y$$ =1. Then either x=1 or if x= negative, y should be positive even integer- Not sufficient since there are two possibilities.

St.2 : Doesn't say anything apart from x is negative-not sufficient.

Combining,

X= negative, and resultant value is 1 then x= -1 , and y should be positive even integer(Multiples of 2).

Then, y is divisible by 2. So the answer is C.
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Re: If x and y are integers, is y divisible by 2? (1) x^y=1 [#permalink]

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09 Jun 2014, 08:52
I solved it the following way:

x,y = int
is y/2=int?

(1)
x^y = 1.

This could happen in two ways.
a) x = 1, y = any number.
b) y = 0, x = anything but 0.
c) x =-1, y = an even number

As y can be any number, I cannot answer this question with the information from statement 1.
Hence insufficient. I am crossing off A and D.

(2)
x= negative.

The information I have right now is that x and y are integers. The fact that x is negative tells me nothing about y, as I haven't been told about any relationship between them.

Insufficient.

(1)+(2)
As I had the previous possibilities written down from (1):
a) x = 1, y = any number.
b) y = 0, x = anything but 0.
c) x=-1, y = any even number

I can conclude that as x is negative(2) the scenarios (a) and (b) are not possible. I am therefore left with scenario c. Which tells me that y is an even number (which is always divisible by 2).

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Re: If x and y are integers, is y divisible by 2? [#permalink]

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09 Jun 2014, 08:54
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Expert's post
If x and y are integers, is y divisible by 2?

(1) $$x^y=1$$. If x=1, then y could be any integer even or odd. Not sufficient.

(2) x is negative. No info about y. Not sufficient.

(1)+(2) x is negative and $$x^y=1$$. We can have the following two cases

i. If $$x=-1$$, then y must be even: $$(-1)^{even}=1$$.
ii. If x is any other negative integer, then y must be 0: $$(non-zero \ integer)^{0}=1$$.

In either case y is even, so divisible by 2. Sufficient.

Hope it's clear.
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Re: If x and y are integers, is y divisible by 2? [#permalink]

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15 Mar 2016, 21:06
goodyear2013 wrote:
If x and y are integers, is y divisible by 2?

(1) $$x^y=1$$
(2) x is negative.

the question really asks if y is even.
i spent some extra time thinking = is 0 divisible by 2? yes, 0 is divisible by 2, as there is no remainder.

1. x^y = 1
1^2 = 1 - yes
1^3 = 1 - no
1 alone is insufficient.

2. x < 0 - not sufficient.

1+2 -> if the base is negative, then the only way it can be positive if the power is even.
thus, it can be x^0=1, x^2 =1, x^4 = 1.
so yes, y is even. and since 0 is divisible by 2, then C is the correct answer.

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Re: If x and y are integers, is y divisible by 2? [#permalink]

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16 Mar 2016, 00:13
Here we need to check whether o not y is even
Now x^y =1 there are three cases which support this statement
for y=0 ; x => anything but zero (y=even)
for y=anything ; x=1 (y can be even or odd)
for x=-1 => y =>even (y is even )
hence B is correct.
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Re: If x and y are integers, is y divisible by 2? [#permalink]

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28 Sep 2017, 08:45
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Re: If x and y are integers, is y divisible by 2?   [#permalink] 28 Sep 2017, 08:45
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