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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
x and y are integers, and x + y > 0
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25 Feb 2020, 12:48
25 Feb 2020, 12:48
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Question Stats:
25% (02:52) correct
75% (05:16) wrong based on 4 sessions
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\(x\) and \(y\) are integers, and \(x + y > 0\). If \(5x - 1\) is a multiple of \(4\), and \(4y + 2\) is a multiple of \(6\), what is the smallest possible value of \(x + y\)?
A) 0
B) 1
C) 3
D) 5
E) 7
A) 0
B) 1
C) 3
D) 5
E) 7
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: x and y are integers, and x + y > 0
[#permalink]
27 Feb 2020, 06:29
27 Feb 2020, 06:29
GreenlightTestPrep wrote:
\(x\) and \(y\) are integers, and \(x + y > 0\). If \(5x - 1\) is a multiple of \(4\), and \(4y + 2\) is a multiple of \(6\), what is the smallest possible value of \(x + y\)?
A) 0
B) 1
C) 3
D) 5
E) 7
A) 0
B) 1
C) 3
D) 5
E) 7
The key here is to recognize that multiples of a value need not necessarily be positive.
For example some multiples of 5 include: 5, -15, 0, 55, -20 etc
So from here it's simply a matter of finding some values that satisfy the given conditions.
If \(5x - 1\) is a multiple of \(4\), then it could be the case that x = -3, since \(5(-3)-1 = -16\) and -16 is a multiple of 4.
Likewise, if \(4y + 2\) is a multiple of \(6\), then it could be the case that y = 4, since \(4(4)+2 = 18\) and 18 is a multiple of 6.
So, x + y = (-3) + 4 = 1
So 1 is the smallest possible value of x + y
Answer: B
ASIDE: The answer is not A, because we are told that \(x + y > 0\).
Cheers,
Brent
gmatclubot
Re: x and y are integers, and x + y > 0 [#permalink]
27 Feb 2020, 06:29
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