2014 AIME I Problems/Problem 3
Problem 3
Find the number of rational numbers such that when 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 to where each each element has but we also need to be irreducible
we note that hence is irreducible iff isn't, which is equivalent to m not being divisible by 2 or 5. so the question is equivalent to "how many numbers between 501 and 999 are not divisible by neither 2 or 5?"
we note there are 499 numbers between 501 and 999
- 249 are even (divisible by 2)
- 99 are divisible by 5
- 49 are divisible by 10 (both 2 and 5)
Using Principle of Inclusion Exclusion (PIE): we get: numbers between 501 and 999 arent divisible by neither 2 or 5 so our answer is