2008 AMC 10A Problem 9

1. The expression has two fractions, so combine them first. Use the common bottom $6$:
$$\frac{2x}{3} - \frac{x}{6} = \frac{4x}{6} - \frac{x}{6} = \frac{3x}{6} = \frac{x}{2}.$$
2. So the whole condition is just "$\frac{x}{2}$ is a whole number". That happens exactly when $x$ is an even whole number.
3. Now test the choices with a small example. Take $x = 2$: then $\frac{x}{2} = 1$, a whole number, so $x = 2$ is allowed.
4. But $2$ is positive, not a multiple of $3$, not a multiple of $6$, and not a multiple of $12$. That rules out (A), (C), (D) and (E).
5. The one thing that is always true is that $x$ is even, and $x = 2$ shows it need not be a multiple of $3$.

The answer is $\boxed{\textbf{(B)}\ \text{It is even, but not necessarily a multiple of 3.}}$

Simplify before you reason. The $3$ and the $6$ in the bottoms look important, but they disappear once the fractions are combined. For a "must be true" question, one small example that satisfies the condition can knock out several choices at once.

Put the two fractions over the common denominator $6$:
$$\frac{2x}{3} - \frac{x}{6} = \frac{4x}{6} - \frac{x}{6} = \frac{3x}{6} = \frac{x}{2}.$$

So the condition is simply that $\frac{x}{2}$ is an integer, which means $x$ is an even integer.

Is $x$ forced to be a multiple of $3$? No: $x = 2$ gives $\frac{x}{2} = 1$, an integer, and $2$ is not a multiple of $3$. Negative values are not forced either ($x = 2$ again). So the only guaranteed property is evenness.

The answer is $\boxed{\textbf{(B)}\ \text{It is even, but not necessarily a multiple of 3.}}$