2014 AIME I Problems/Problem 3

Problem 3

Find the number of rational numbers $r,$ $0<r<1,$ such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution

We have that the set of these rational numbers is from $\dfrac{1}{999}$ to $\dfrac{499}{501}$ where each each element $\dfrac{n}{m}$ has $n+m =1000$ and $\dfrac{n}{m}$ is irreducible.

We note that $\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1$ . Hence, $\dfrac{n}{m}$ is irreducible if $\dfrac{1000}{m}$ is irreducible, and $\dfrac{1000}{m}$ is irreducible if $m$ is not divisible by 2 or 5. Thus, the answer to the question is the number of integers between 999 and 501 inclusive that are not divisible by 2 or 5.

We note there are 499 numbers between 501 and 999, and

249 are divisible by 2
99 are divisible by 5
49 are divisible by 10

Using the Principle of Inclusion and Exclusion, we get that there are $499-249-99+49=200$ numbers between $501$ and $999$ are not divisible by either $2$ or $5$ , so our answer is $\boxed{200}$ .

Euler's Totient Function can also be used to arrive at 400 numbers relatively prime to 1000, meaning 200 possible fractions satisfying the necessary conditions.

2014 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2014 AIME I Problems/Problem 3

Problem 3

Solution

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