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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 (squared)+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) is z is even, then xy is even

e) xy is even

### 2 Answers by Expert Tutors

Linghong C. | Experienced Math tutorExperienced Math tutor
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odd+odd=even  odd+even=odd, even +even =even

odd*odd=odd, odd*even=even, even*even=even

(odd)2=odd, (even)2= even

x+y is even, hence, either x and y are both even or x and y are both odd. Thus x can be either even or odd.    a) and b) are both wrong.

When x and y are both even, xy is even; when both x and y are odd, xy is odd. Hence,  xy can be either even or odd.  e) is wrong

Since x+y is even,  (x+y)2 is also even.  (x+y)+x+z is odd.  Hence, even+(x+z)=odd . x+z must be odd. Hence when z is even,  x  must be odd.   c) is correct.

Since x and y must be either both odd or both even, and we also know when z is even, x must be odd. Hence,when z is even,  y must be odd too.  xy is odd.   d) is wrong.