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For all integers a and b, the operation θ is defined by aθb = (a+b)/2.

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Bunuel
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For all integers a and b, the operation θ is defined by aθb = (a+b)/2. [#permalink] For all integers a and b, the operation θ is defined by aθb = (a+b)/2.   23 May 2020, 04:10
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For all integers a and b, the operation θ is defined by \(aθb = \frac{a + b}{2}\). If x and y are integers, is \(x θ y\) an integer?


(1) \(x^2 = y^2\)

(2) \(x = –y\)


DS20593
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D
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Re: For all integers a and b, the operation θ is defined by aθb = (a+b)/2. [#permalink] For all integers a and b, the operation θ is defined by aθb = (a+b)/2.   23 May 2020, 05:58
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Bunuel wrote:
For all integers a and b, the operation θ is defined by \(θ = \frac{a + b}{2}\). If x and y are integers, is \(x θ y\) an integer?


(1) \(x^2 = y^2\)

(2) \(x = –y\)


PS20593




xθy = (x + y)/2, which will be an integer as long as x + y is even.

If x^2 = y^2, then either x = y, in which case x + y = x + x = 2x, which is even, or -x = y, in which case x + y = x - x = 0, which is again even. So Statement 1 guarantees that x + y is even and is sufficient.

Statement 2 is a special case of Statement 1; it tells us that xθy = 0, so is also sufficient, and the answer is D.

I think there's a typo in the question, though -- the operation is not defined properly, as written. It should say:

"the operation θ is defined by \(aθb = \frac{a + b}{2}\)"
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Re: For all integers a and b, the operation θ is defined by aθb = (a+b)/2. [#permalink] For all integers a and b, the operation θ is defined by aθb = (a+b)/2.   25 May 2020, 23:48
For all integers a and b, the operation θ is defined by aθb=a+b2aθb=a+b2. If x and y are integers, is xθyxθy an integer?


(1) x^2=y^2

(2) x=–y

The average of 2 numbers will be an integer if only the sum of the two numbers is divisible by 2.
statement 1 : x^2 = y^2
x=2 y=2 (x+y)/2 =2=integer
x=-2 y=-2 (x+y)/2 =0= integer Suff

statement 2 :
(2) x=–y
its a subset of statement 1
Hence Suff

Answer D
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For all integers a and b, the operation θ is defined by aθb = (a+b)/2. [#permalink] For all integers a and b, the operation θ is defined by aθb = (a+b)/2.   21 Jun 2020, 15:20
Bunuel wrote:
For all integers a and b, the operation θ is defined by \(aθb = \frac{a + b}{2}\). If x and y are integers, is \(x θ y\) an integer?


(1) \(x^2 = y^2\)

(2) \(x = –y\)


DS20593



Can someone find my flaw in thinking please?

Let us assume 1: in such case, x could be 1 and y -1, still satisfying (1). In such case, original question is not an integer. However, if we assume 1/1, then original question is an integer =1.


EDIT: DISREGARD. My flaw was in the assumption that a neutral integer would make the original question wrong lol. 0 still satisfies. silly me!
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Re: For all integers a and b, the operation θ is defined by aθb = (a+b)/2. [#permalink] For all integers a and b, the operation θ is defined by aθb = (a+b)/2.   25 Jun 2020, 20:21
Bunuel wrote:
For all integers a and b, the operation θ is defined by \(aθb = \frac{a + b}{2}\). If x and y are integers, is \(x θ y\) an integer?


(1) \(x^2 = y^2\)

(2) \(x = –y\)



Since only even numbers are divisible by 2, this question can be rephrased as "is x+y even?". Remember that zero is an even number.

S1: x and y can be zero, the same number with the same signs or the same number with opposite signs. Regardless, x+y will always be zero or another even number. SUFFICIENT.

S2: This is only true x and y = 0. SUFFICIENT.

ANSWER: D
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Re: For all integers a and b, the operation θ is defined by aθb = (a+b)/2. [#permalink] For all integers a and b, the operation θ is defined by aθb = (a+b)/2.   27 Jul 2020, 19:47
(x+y)/2 = int?

Given that x and y are integers, the only thing that won't make x+y/2 an integer is if x+y is odd

1. x^2 = y^2

Irrespective of whether x is positive or negative or whether y is positive or negative, we know that their squares equal each other, so worst case:
x+y = 0 i.e. even therefore 0/2 = 0 -->true

any number will work since we know x must equal at least the negative equivalent of y, therefore either Odd+ odd = even or even + even = even

sufficient

2. same logic
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Re: For all integers a and b, the operation is defined by ab = (a+b)/2. [#permalink] For all integers a and b, the operation θ is defined by aθb = (a+b)/2.   24 Oct 2023, 01:33
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Re: For all integers a and b, the operation is defined by ab = (a+b)/2. [#permalink]
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