AMC 10 · 2008 · #9

Grade 7 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

fraction-arithmeticparitydivisibility-rules guess-and-checklogical-deduction ↑ Prerequisites: fraction-arithmetic

📏 Medium solution 💡 2 insights

Problem

x is an integer, and the expression

\frac{2x}{3}-\frac{x}{6}

is also an integer. Which one of the five descriptions of x must always be true?

Pick an answer.

(A)

$\ \text{It is negative.}$

(B)

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

(C)

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

(D)

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

(E)

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

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The messy-looking expression hides a much simpler one. Split the task into two smaller problems: first simplify 2x/3-x/6 to a single fraction, then ask what that fraction being an integer demands of x. Once the demand is clear, test a value to see which extra properties are NOT forced, and eliminate the choices that claim too much.

1STEP 1

Combine into one fraction

Over the common denominator 6, $\frac{2x}{3}=\frac{4x}{6}$ , so the difference is $\frac{3x}{6}$ , which reduces to $\frac{x}{2}$ .

2x/3-x/6=4x/6-x/6=3x/6=x/2

2STEP 2

Translate the integer condition

So the given condition just says $\frac{x}{2}$ is a whole number, which happens exactly when 2 divides x.

x/2inZ⇔ 2 ∣ x

3STEP 3

Test what is not forced

Test x=2: $\frac{2}{2}=1$ is an integer, yet 2 is positive and is not a multiple of 3.

x=2: 2/2=1inZ, 2=2·1, 2 ≠ 3k

4STEP 4

Eliminate and conclude

That one example kills negative, multiple of 3, multiple of 6, and multiple of 12, so only even survives — choice (B).

x even, x not necessarily a multiple of 3

Answer

It is even, but not necessarily a multiple of 3.

Check both directions. If x is even, write x=2k; then x/2=k is an integer, so every even x works. If x is odd, x/2 ends in .5 and is never an integer. So the allowed x are precisely the even numbers - no more, no less - which matches choice (B).

💡Key takeaway

Simplify the messy expression first: 2x/3 - x/6 is really just x/2, so x only has to be even.

Combine into one fraction
Translate the integer condition
Test what is not forced
Eliminate and conclude

Combine into one fraction

Translate the integer condition

Test what is not forced

Eliminate and conclude

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