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x=2y-7+3z, which of the following must be odd?

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Concentration: Accounting, Finance
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x=2y-7+3z, which of the following must be odd?  [#permalink]

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New post 26 Jan 2018, 15:41
2
1
00:00
A
B
C
D
E

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  75% (hard)

Question Stats:

48% (01:21) correct 52% (01:35) wrong based on 71 sessions

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If x, y , z are integers and x=2y-7+3z, which of the following must be odd?

a) y
b) z
c) xy-1
d) xz-1
e) none of these
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Re: x=2y-7+3z, which of the following must be odd?  [#permalink]

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New post 26 Jan 2018, 17:25
2
3
selim wrote:
If x, y , z are integers and x=2y-7+3z, which of the following must be odd?

a) y
b) z
c) xy-1
d) xz-1
e) none of these


x=2y-7+3z......
y can be anything but 2y-7 will be even-odd=odd..
So x=odd+3z...x-3z=0dd..
Therefore one of x and z is odd and other even...
xz=O*E=even..... and xz-1= even-1=odd..
D
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Re: x=2y-7+3z, which of the following must be odd?  [#permalink]

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New post 27 Jan 2018, 04:57
1
Given : x= 2y-7 +3z
With this given equation, we do not know y or z must be odd. It can be even or odd. So, option A and B is not correct.
Next ,
2y - always even , 2y - 7= Even - Odd = Odd
Now, x = Odd + 3z
Here, if z is even , x= Odd + Even = Odd
If z is odd , x= Odd + Odd = Even
In either case, xz = O*E or E* O = Even and xz-1 = Even -1 must always be odd.
Hence , option D)
Re: x=2y-7+3z, which of the following must be odd? &nbs [#permalink] 27 Jan 2018, 04:57
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