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# If x and y are integers, is x even?

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If x and y are integers, is x even?  [#permalink]

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26 Sep 2016, 03:45
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55% (hard)

Question Stats:

63% (01:40) correct 37% (01:49) wrong based on 164 sessions

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If x and y are integers, is x even?

(1) x+y=y^5

(2) x+y=3y

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If x and y are integers, is x even?  [#permalink]

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26 Sep 2016, 06:11
Bunuel wrote:
If x and y are integers, is x even?

(1) x+y=y^5

(2) x+y=3y

Stat 1: odd + even = even ...not this..we have to get odd...
or even+even = even ...this can be both y and y^5 integer type is same.
or odd+even = even...not this
or even + odd = odd...this can be both y and y^5 integer type is same...x can be even...Sufficient..

Stat 2: x = 2y...whatever the value of y be...x will always even..since y is multiplied by 2.

x is even...

IMO option D.
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Re: If x and y are integers, is x even?  [#permalink]

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26 Sep 2016, 06:27
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Bunuel wrote:
If x and y are integers, is x even?

(1) x + y = y⁵

(2) x + y = 3y

Some important rules:
1. ODD +/- ODD = EVEN
2. ODD +/- EVEN = ODD
3. EVEN +/- EVEN = EVEN

4. (ODD)(ODD) = ODD
5. (ODD)(EVEN) = EVEN
6. (EVEN)(EVEN) = EVEN

Target question: Is x even?

Given: x and y are integers

Statement 1: x + y = y⁵
Subtract y from both sides to get: x = y⁵ - y
Let's examine the two possible cases: y is EVEN and y is ODD
case a: y is ODD. So, x = y⁵ - y = ODD⁵ - ODD = ODD - ODD = EVEN
case b: y is EVEN. So, x = y⁵ - y = EVEN⁵ - EVEN = EVEN - EVEN = EVEN
In both possible cases, x is even. So, x MUST be even.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x + y = 3y
Subtract y from both sides to get: x = 2y
If x equals the product of 2 and some integer, then we can be certain that x is even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

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Re: If x and y are integers, is x even?  [#permalink]

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26 Sep 2016, 09:49
Bunuel wrote:
If x and y are integers, is x even?

(1) x+y=y^5

(2) x+y=3y

Statement 1: Sufficient

x = y^5-y => x= y(y^4-1)

Assuming y as even : y^4 - 1 = EVEN - ODD = ODD
y(y^4-1) = EVEN*ODD =EVEN therefore x is even
Assuming y as odd : y^4 - 1 = ODD - ODD = EVEN
y(y^4-1) = ODD*EVEN =EVEN therefore x is even

Statement 2: Sufficient

x+y=3y can be re-written as x = 2y => x is divisible by 2 therefore x is even

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Re: If x and y are integers, is x even?  [#permalink]

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27 Sep 2016, 07:58
statement 1 = x = y^5 - y

y can be odd or even but the result of subtracting 2 odds or 2 evens is always even

statement 2 : x = 2y clearly sufficient

Choice D
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Re: If x and y are integers, is x even?  [#permalink]

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27 Sep 2016, 08:20
(1) x+y=y^5

x= y ($$y^4$$-1)

If y is even, x is even.
If y is odd. ($$y^4$$-1) will be even. ==> x is even

--Sufficient.

(2) x+y=3y

x=2y. x is even. --Sufficient.

Ans. D.
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Re: If x and y are integers, is x even?  [#permalink]

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19 Nov 2016, 08:08
Here we need to get whether x is even or not
Statement 1
x+y=y^5
hence x=y^5-y
here if y=odd
x=odd-odd=even
if y=even
x=even-even=even
Hence x is always even
hence sufficient
Statement 2
x+y=3y
x=2y
as y is an integer => 2y is always even
Hence Sufficient
Hence D
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Re: If x and y are integers, is x even?  [#permalink]

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19 Aug 2018, 12:26
(1) x=y^5-y=y(y^4-1)=y(y^2+1)(y^2-1)=y(y^2+1)(y+1)(y-1) <-- 3 consecutive integers. It's even
(2) x=2y is even
Re: If x and y are integers, is x even?   [#permalink] 19 Aug 2018, 12:26
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