AMC 10 Step by Step
Log inSign up
Printable

2008 AMC 10A Problem 9

Problem onlyBack to the problem
2008 AMC 10A Problem 9AMC 10 Step by Step
Name:                             Date: September 13, 2026
Divisibility & Factors · difficulty 2 of 5 · about 2 min · position P1-10

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

  1. A)

    It is negative.\text{It is negative.}

  2. B)

    It is even, but not necessarily a multiple of 3.\text{It is even, but not necessarily a multiple of 3.}

  3. C)

    It is a multiple of 3, but not necessarily even.\text{It is a multiple of 3, but not necessarily even.}

  4. D)

    It is a multiple of 6, but not necessarily a multiple of 12.\text{It is a multiple of 6, but not necessarily a multiple of 12.}

  5. E)

    It is a multiple of 12.\text{It is a multiple of 12.}

Answer

B

Key insight

Combine the fractions: 2x/3 - x/6 = x/2, so x is an even integer and nothing more; x = 2 rules out any multiple-of-3 claim.

Solution

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.}}

Why this works

Simplify before reasoning: the 33 and 66 in the denominators are a distraction that vanishes once the fractions are combined. A "must be true" question has two halves: prove the property that always holds (xx even) and produce one counterexample for anything stronger (x=2x = 2).

The trap

Assuming the denominators 3 and 6 force x to be a multiple of 3 or 6 without simplifying the expression first.

Common mistakes

  • Assuming the denominators 3 and 6 force x to be a multiple of 3 or 6 without simplifying the expression first.
  • Picking (D) or (E) after testing only convenient values like x=6x = 6 or x=12x = 12, which satisfy the condition but do not show it is necessary.

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer