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
and
is irreducible.
We note that .
Hence,
is irreducible if
is irreducible, and
is irreducible if
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 numbers between
and
are not divisible by either
or
, so our answer is
.
Euler's Toingent Function can also be used to arrive at 400 numbers relatively prime to 1000, meaning 200 possible fractions satisfying the necessary conditions.
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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