x, y and z are consecutive positive integers such that x < y < z; which of the following must be true?
1. xyz is divisible by 6
2. (z-x)(y-x+2) = 4
3. xy is odd
Choose the correct choice from the following answers
A) I only
B) II only
C) III only
D) I 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 would be true as at least one of the three numbers is divisible by 2 and one by 3 so xyz would be divisible by 6.   Statement 2 can be simplified if we write everything in terms of y as ((y+1)-(y-1))*(y-(y-1)+2) = 2*3= 
<gadget id="calculator">2*3</gadget>
<output>6</output>
6 So, not 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 is always true and hence answer is   I only.
<result>D</result>
x, y and z are consecutive positive integers such that x < y < z; which of the following must be true?
1. xyz is divisible by 6
2. (z-x)(y-x+2) = 4
3. xy is odd