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GRE Math Challenge #97-w, x, y, and z are consecutive positi [#permalink]
Expert's post 00:00

Question Stats: 46% (00:38) correct 53% (00:59) wrong based on 26 sessions
w, x, y, and z are consecutive positive integers and w < x < y < z.

Quantity A: The remainder when (w +x)(x + y)(y + z) is divided by 2
Quantity B: 1

• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
[Reveal] Spoiler: OA

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Sandy
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Re: GRE Math Challenge #97 [#permalink]
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Expert's post
sandy wrote:
w, x, y, and z are consecutive positive integers and w < x < y < z.

Quantity A: The remainder when (w +x)(x + y)(y + z) is divided by 2
Quantity B: 1

IMPORTANT CONCEPT: consecutive integers alternate from ODD to EVEN
So, the sum of any two consecutive integers will always be ODD, since we're invariably adding an ODD integer and an EVEN integer.

Since w < x < y < z, we can conclude that:
w and x are consecutive integers, which means w + x is ODD
x and y are consecutive integers, which means x + y is ODD
y and z are consecutive integers, which means y + z is ODD

So, (w +x)(x + y)(y + z) = (ODD integer)(ODD integer)(ODD integer)
= ODD integer

If we divide any odd integer by 2, the remainder will be 1.

So, we get:
Quantity A: The remainder when (w +x)(x + y)(y + z) is divided by 2 = 1
Quantity B: 1

[Reveal] Spoiler:
C

Cheers,
Brent
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Re: GRE Math Challenge #97-w, x, y, and z are consecutive positi [#permalink]
i did it with simple logic of numbers and remainder turned out to be 0
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Re: GRE Math Challenge #97-w, x, y, and z are consecutive positi [#permalink]
my mistake even if we plug in simple nos we still get remainder as 1..hence C is correct
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Re: GRE Math Challenge #97-w, x, y, and z are consecutive positi [#permalink]
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Re: GRE Math Challenge #97-w, x, y, and z are consecutive positi [#permalink]
If you don't see that every sum within the brackets yields an ODD integer just try plugging in numbers (it's important however to figure out that ANY odd number divided by 2 will yield a remainder of 1)
Use the most easiest one, hence w = 1, x = 2, y = 3 and z = 4. Plugging in gives 105 (ODD) -> remainder will be 1
To confirm the answer try another set of numbers: w = 2, x = 3, y = 4 and z = 5. Plugging in, again gives an odd number 315 -> remainder 1
At this point you should realize that any combination of consecutive integers in the given equation (w+x)*(x+y)*(y+z) will yield an odd number and therefore always leave us with a remainder of 1 when divided by 2. -> Answer C Re: GRE Math Challenge #97-w, x, y, and z are consecutive positi   [#permalink] 25 Jan 2018, 02:08
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