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GRE Math Challenge #97-w, x, y, and z are consecutive positi
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Updated on: 20 Feb 2020, 10:42
Updated on: 20 Feb 2020, 10:42
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49% (01:00) wrong based on 79 sessions
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w, x, y, and z are consecutive positive integers and \(w < x < y < z\).
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.
Kudos for the right answer and explanation
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.
Kudos for the right answer and explanation
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: GRE Math Challenge #97
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17 May 2015, 16:46
17 May 2015, 16:46
2
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
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
Answer:
Show: ::
C
Cheers,
Brent
Re: GRE Math Challenge #97-w, x, y, and z are consecutive positi
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22 Jan 2018, 07:17
22 Jan 2018, 07:17
i did it with simple logic of numbers and remainder turned out to be 0
