Question 92788: If x, y, and z are positive intergers such that the value of x+y is even and the value of +x+z is odd, which of the following must be true?
(A) x is odd
(B) x is even
(C) If z is even, then x is odd
(D) If z is even, then xy is even
(E) xy is even
Answer by Edwin McCravy(20060) (Show Source):
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If x, y, and z are positive integers such that the value of x+y
is even and the value of (x+y)²+x+z is odd, which of the following
must be true?
(A) x is odd
(B) x is even
(C) If z is even, then x is odd
(D) If z is even, then xy is even
(E) xy is even
There are 8 possible cases:
We can use any even and odd integer. The easiest even positive
integer is 2 and the easiest odd integer is 1. So let's use
these in those 8 cases
x y z x+z=? (x+y)²+x+z=?
------------------------------------------------------------
1. 2 2 2 2+2=4=EVEN (2+2)²+2+2=4²+2+2=16+2+2=20=EVEN
2. 2 2 1 2+1=3=ODD (2+2)²+2+1=4²+2+1=16+2+1=19=ODD
3. 2 1 2 2+2=4=EVEN (2+1)²+2+2=3²+2+2= 9+2+2=13=EVEN
4. 2 1 1 2+1=3=ODD (2+1)²+2+1=3²+2+1= 9+2+1=12=EVEN
5. 1 2 2 1+2=3=ODD (1+2)²+1+2=3²+1+2= 9+1+2=12=EVEN
6. 1 2 1 1+2=3=ODD (1+2)²+1+1=3²+1+1= 9+1+1=11=ODD
7. 1 1 2 1+1=2=EVEN (1+1)²+1+2=2²+1+2= 4+1+2= 7=ODD
8. 1 1 1 1+1=2=EVEN (1+1)²+1+1=2²+1+1= 4+1+1= 6=EVEN
There are only one case when x+y is EVEN and (x+y)²+x+z is ODD,
That is case 7, where x=1, y=1 and z=2
So x is odd, y is odd, and z is even.
The only one of the choices that must be true is A.
Edwin
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