# Practice GMAT Problem Solving Question

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+1)=4
3. xy is odd
 A) I only B) II only C) III only D) I and II only E) I, II, and III
1. Evaluate each statement one by one, starting with the first.
2. Evaluate Statement I.
1. In order to be divisible by 6, a number must have the prime factors of 6, which are 2 and 3.
2. In a set of 3 consecutive integers, at least one of them will be even. Any even number has a 2 as one of its prime factors. Thus, the product of the 3 consecutive positive integers will have a factor of 2.
3. Any consecutive series of 3 integers has a multiple of 3 in it, since every third integer is a multiple of 3. Thus, either x, y, or z is a multiple of 3 and therefore has 3 as one of its prime factors. Thus, the product xyz will have a factor of 3.
4. Since xyz will have a 3 and a 2 in its prime factorization tree, xyz must be divisible by 6. Therefore, I is always true.
3. Evaluate Statement II.
1. Since x, y, and z are consecutive integers such that x < y < z, we can rewrite y and z in terms of x: y=x+1, and z=x+2.
2. Substitute these values in the equation:
(z-x)(y-x+1)=4
([x+2]-x)([x+1]-x+1)=4
3. Simplify the equation:
(x-x+2)([x-x+1+1)=4
(2)(1+1) = 2(2) = 4
4. It is clear that II will always be true.
4. Evaluate Statement III.
1. Since x, y, and z are consecutive numbers such that x < y < z, if y is even, then x and z will both be odd. The product of two odd numbers is odd so xz would be odd in this case. But, xy = (odd)(even) = even.
2. But, if y is odd, then x and z will be even. The product of two even numbers is even, so xz is even in this case and xy is also even.
3. Since there is no way to guarantee that both x and y are odd, we cannot conclude that statement III is always true.
4. Note: Since x and y are consecutive integers, either x or y will always be even. Consequently, xy will always be even: either (even)(odd) = even OR (odd)(even) = even.
5. Since I and II must be true, but III is not always true, the correct answer is D.