x, y and z are consecutive positive integers such that x < y < z; which of the following must be true?
1. yz is even
2. (z-x)(y-x+1) = 4
3. xy is odd  Select:
A) I only	B) I , II only	C) III only	D) I and II only	E) I, II, and III
Since x, y and z are consecutive integers such that x &lt; y &lt; z, we can say x = y-1 and Z = y+1   Statement 1 yz =y*(y+1) ..if one is odd then other will be even and vice versa ..true   Statement 2 can be simplified if we write everything in terms of y as ((y+1)-(y-1))*(y-(y-1)+1) = 2*2= 
<gadget id="calculator">2*2</gadget>
<output>4</output>
4 So, always true   Statement 3 talks about xy Since x and y are consecutive integers one of them is odd and other is even so product would always be even and hence not true.   So, I and II are always true and hence answer is   I , II only
<result>B</result>
x, y and z are consecutive positive integers such that x < y < z; which of the following must be true?
1. yz is even
2. (z-x)(y-x+1) = 4
3. xy is odd