SOLUTION: If x, y, and z are positive intergers 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 e

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): You can put this solution on YOUR website!


If x, y, and z are positive integers such that the value of x+y
is even and the value of (x+y)�+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)�+x+z=?
------------------------------------------------------------  
1.   2  2  2   2+2=4=EVEN   (2+2)�+2+2=4�+2+2=16+2+2=20=EVEN 
2.   2  2  1   2+1=3=ODD    (2+2)�+2+1=4�+2+1=16+2+1=19=ODD   
3.   2  1  2   2+2=4=EVEN   (2+1)�+2+2=3�+2+2= 9+2+2=13=EVEN
4.   2  1  1   2+1=3=ODD    (2+1)�+2+1=3�+2+1= 9+2+1=12=EVEN
5.   1  2  2   1+2=3=ODD    (1+2)�+1+2=3�+1+2= 9+1+2=12=EVEN 
6.   1  2  1   1+2=3=ODD    (1+2)�+1+1=3�+1+1= 9+1+1=11=ODD
7.   1  1  2   1+1=2=EVEN   (1+1)�+1+2=2�+1+2= 4+1+2= 7=ODD
8.   1  1  1   1+1=2=EVEN   (1+1)�+1+1=2�+1+1= 4+1+1= 6=EVEN

There are only one case when x+y is EVEN and (x+y)�+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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