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D
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Difficulty:
55%
(hard)
Question Stats:
65% (02:18) correct
35%
(02:24)
wrong
based on 331
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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. y + zx
A. y
B. z
C. xy - 1
D. xz - 1
E. y + zx
26 Jan 2018, 17:25
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selim
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
27 Jan 2018, 04:57
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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)
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)
