If x, y, and z are positive integers such that the value of.... PLEASE HELP

Question

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

Answer 1

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.

 

Answer 2

Because x + y is even, either x and y are both odd or x and y are both even.

Since x + y is even, (x + y)² must also be even.  Since (x + y)² + x + z is odd, then either x or z is odd, but not both.

C is the correct answer.