2008 AMC 10A Problem 9
Suppose that is an integer. Which of the following statements must be true about ?
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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 :
2. So the whole condition is just " is a whole number". That happens exactly when is an even whole number.
3. Now test the choices with a small example. Take : then , a whole number, so is allowed.
4. But is positive, not a multiple of , not a multiple of , and not a multiple of . That rules out (A), (C), (D) and (E).
5. The one thing that is always true is that is even, and shows it need not be a multiple of .
The answer is
Why this works
Simplify before you reason. The and the 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 :
So the condition is simply that is an integer, which means is an even integer.
Is forced to be a multiple of ? No: gives , an integer, and is not a multiple of . Negative values are not forced either ( again). So the only guaranteed property is evenness.
The answer is
Review in the AoPS books
- Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
- Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)
- Introduction to Algebra, Chapter 20: Special Functions (§20.3)
- Introduction to Number Theory, Chapter 13: Divisibility Rules (§13.2)
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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.