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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]
Answer is C.
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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
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Re: GRE Math Challenge #97-w, x, y, and z are consecutive positi [#permalink]
1
sandy wrote:
w, x, y, and z are consecutive positive integers and \(w < x < y < z\).


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


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

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As w,x,y,z are consecutive positive integers, they will be no fraction. Let's assume that these numbers are 2,3,4,5 and according to this \([w +x][x + y][y + z]\) summation is = 5*7*9 and the result is = odd.
Now let's take other consecutive numbers, 5,6,7,8 and summation is = 11*13*15 and the result is odd.

If an odd number is going to be divided by 2, the remainder will be 1 always. That's why the answer is (C).
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Re: GRE Math Challenge #97-w, x, y, and z are consecutive positi [#permalink]
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