AMC 10 Step by Step

2008 AMC 10A Problem 9

P1-10~2 minPrint

Suppose that 2x3−x6\frac{2x}{3}-\frac{x}{6} is an integer. Which of the following statements must be true about xx ?

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Key insight

Over a common bottom of 6 the expression becomes x over 2. So x must be even, and x = 2 shows it need not be a multiple of 3.

Solution

1. The expression has two fractions, so combine them first. Use the common bottom 66:
2x3−x6=4x6−x6=3x6=x2.\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 "x2\frac{x}{2} is a whole number". That happens exactly when xx is an even whole number.
3. Now test the choices with a small example. Take x=2x = 2: then x2=1\frac{x}{2} = 1, a whole number, so x=2x = 2 is allowed.
4. But 22 is positive, not a multiple of 33, not a multiple of 66, and not a multiple of 1212. That rules out (A), (C), (D) and (E).
5. The one thing that is always true is that xx is even, and x=2x = 2 shows it need not be a multiple of 33.

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

Why this works

Simplify before you reason. The 33 and the 66 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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Put the two fractions over the common denominator 66:
2x3−x6=4x6−x6=3x6=x2.\frac{2x}{3} - \frac{x}{6} = \frac{4x}{6} - \frac{x}{6} = \frac{3x}{6} = \frac{x}{2}.

So the condition is simply that x2\frac{x}{2} is an integer, which means xx is an even integer.

Is xx forced to be a multiple of 33? No: x=2x = 2 gives x2=1\frac{x}{2} = 1, an integer, and 22 is not a multiple of 33. Negative values are not forced either (x=2x = 2 again). So the only guaranteed property is evenness.

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.