AMC 10 · 2008 · #5

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
An expression built from two fractions of a number turns out to be a whole number. Decide which description of that number is forced.

Pick an answer.

(A)
It is negative.
(B)
It is even, but not necessarily a multiple of 3.
(C)
It is a multiple of 3, but not necessarily even.
(D)
It is a multiple of 6, but not necessarily a multiple of 12.
(E)
It is a multiple of 12.
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

Combining and reducing leaves a much smaller denominator.

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

Translate the integer condition

So the condition says only that the number is even.

x/2inZ⇔ 2 ∣ x
3STEP 3

Test what is not forced

A small test value shows nothing stronger is forced.

x=2: 2/2=1inZ, 2=2·1, 2 ≠ 3k
4STEP 4

Eliminate and conclude

So the answer is even but not necessarily a multiple of three, choice (D).

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

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